leecode更新
This commit is contained in:
markilue
parent
2aaa3a3931
commit
d0a8cfc1cf
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@ -107,6 +107,13 @@
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<version>1.2.0</version>
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</dependency>
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<dependency>
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<groupId>junit</groupId>
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<artifactId>junit</artifactId>
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<version>4.13.2</version>
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<scope>compile</scope>
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</dependency>
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</dependencies>
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@ -0,0 +1,39 @@
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package com.atguigu.gmall.realtime;
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import org.junit.Test;
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/**
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*@BelongsProject: rt-gmall-parent
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*@BelongsPackage: com.atguigu.gmall.realtime
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*@Author: markilue
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*@CreateTime: 2023-05-06 21:14
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*@Description: TODO
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*@Version: 1.0
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*/
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public class Test1 {
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@Test
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public void test() {
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String sinTable = "base_trademark";
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String sinkPK = "id";
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String sinkColumn = "id,tm_name";
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String sinExtend = "";
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checkTable(sinTable,sinkPK,sinkColumn,sinExtend);
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}
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private void checkTable(String sinkTable, String sinkPk, String sinkColumns, String sinkExtend) {
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StringBuilder sql = new StringBuilder();
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sql.append("create table if not exist " + sinkTable + "(");
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String[] columns = sinkColumns.split(",");
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for (String column : columns) {
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sql.append(column + " String,");
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}
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sql.append("PRIMARY KEY (" + sinkPk + "))");
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System.out.println(sql.toString());
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//create table if not exist base_trademark(id String,tm_name String,PRIMARY KEY (id))
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}
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}
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@ -111,5 +111,14 @@ public class TableProcessFunction extends BroadcastProcessFunction<JSONObject, S
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//处理配置数据的时候 提前建立维度表 create table if not exist 表空间.表名(字段名 数据类型)
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private void checkTable(String sinkTable, String sinkPk, String sinkColumns, String sinkExtend) {
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StringBuilder sql = new StringBuilder();
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sql.append("create table if not exist " + sinkTable + "(");
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String[] columns = sinkColumns.split(",");
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for (String column : columns) {
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sql.append(column + " String,");
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}
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sql.append("PRIMARY KEY (" + sinkPk + "))");
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}
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}
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@ -0,0 +1,25 @@
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package com.atguigu.gmall.realtime.utils;
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import java.sql.Connection;
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import java.sql.DriverManager;
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import java.sql.PreparedStatement;
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import java.sql.SQLException;
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/**
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*@BelongsProject: rt-gmall-parent
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*@BelongsPackage: com.atguigu.gmall.realtime.utils
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*@Author: markilue
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*@CreateTime: 2023-05-06 20:57
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*@Description: TODO 编写Phoenix工具类,执行sql
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*@Version: 1.0
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*/
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public class MyPhoenixUtils {
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private static final String URL ="jdbc:phoenix:thin:url=http://Ding202:8765;serialization=PROTOBUF";
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public void executeSQL(String sql) throws SQLException {
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Connection connection = DriverManager.getConnection(URL);
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PreparedStatement preparedStatement = connection.prepareStatement(sql);
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preparedStatement.execute();
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}
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}
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@ -21,7 +21,7 @@ public class LC_1094_CarPooling {
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@Test
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public void test() {
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int[][] trips = {{2, 1, 5}, {
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3, 5, 7
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3, 3, 7
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}};
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int capacity = 4;
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System.out.println(carPooling(trips, capacity));
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@ -72,4 +72,46 @@ public class LC_1094_CarPooling {
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return true;
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}
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//二刷
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public boolean carPooling1(int[][] trips, int capacity) {
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//本质上就是判断某一个位置上是否超过capacity
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//那么考虑把每一个位置上的数算出来
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//由于不知道到底有哪些位置，所以先获取最大值和最小值,构造差分数组
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int min = Integer.MAX_VALUE;
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int max = Integer.MIN_VALUE;
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for (int[] trip : trips) {
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if (min > trip[1]) {
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min = trip[1];
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}
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if (max < trip[2]) {
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max = trip[2];
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}
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}
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//通过最大值和最小值构造差分数组
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int[] diff = new int[max - min + 1];
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for (int[] trip : trips) {
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diff[trip[1] - min] += trip[0];
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diff[trip[2] - min] -= trip[0];//这里之所以不用+1，是因为题意是当时就下去了
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}
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if (diff[0] > capacity) {
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return false;
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}
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//遍历diff还原当前位置
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for (int i = 1; i < diff.length; i++) {
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diff[i] += diff[i - 1];
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if (diff[i] > capacity) {
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return false;
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}
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}
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return true;
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}
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}
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@ -0,0 +1,233 @@
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package com.markilue.leecode.hot100.interviewHot.window;
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import org.junit.Test;
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import java.util.HashMap;
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/**
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*@BelongsProject: Leecode
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*@BelongsPackage: com.markilue.leecode.hot100.second
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*@Author: markilue
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*@CreateTime: 2023-04-18 11:00
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*@Description:
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* TODO 力扣76 最小覆盖子串:
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* 给你一个字符串 s 、一个字符串 t 。返回 s 中涵盖 t 所有字符的最小子串。如果 s 中不存在涵盖 t 所有字符的子串，则返回空字符串 "" 。
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*@Version: 1.0
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*/
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public class LC_76_MinWindow {
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@Test
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public void test() {
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String s = "aa", t = "aa";
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System.out.println(minWindow1(s, t));
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}
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@Test
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public void test1() {
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String s = "ADOBECODEBANC", t = "ABC";
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System.out.println(minWindow2(s, t));
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}
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/**
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* 滑动窗口法
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* 维护一个需要的窗口 和一个 滑动的窗口
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* @param s
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* @param t
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* @return
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*/
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public String minWindow(String s, String t) {
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HashMap<Character, Integer> need = new HashMap<>();
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for (char c : t.toCharArray()) {
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need.put(c, need.getOrDefault(c, 0) + 1);
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}//记录需要的次数
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HashMap<Character, Integer> window = new HashMap<>();
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int left = 0;
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int right = 0;
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int count = 0;//记录满足要求的个数
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int start = 0;
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int len = Integer.MAX_VALUE;
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while (right < s.length()) {
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char c = s.charAt(right++);
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if (need.containsKey(c)) {
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window.put(c, window.getOrDefault(c, 0) + 1);
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if (need.get(c).equals(window.get(c))) {
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count++;
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}
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}
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//是否满足要求了
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while (count == need.size()) {
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if (len > right - left + 1) {
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start = left;
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len = right - left + 1;
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}
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char c1 = s.charAt(left++);
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if (need.containsKey(c1)) {
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Integer num = window.get(c1);
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if (num.equals(need.get(c1))) {
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count--;
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}
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window.put(c1, window.getOrDefault(c1, 0) - 1);//增加窗口中的个数
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}
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}
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}
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return len == Integer.MAX_VALUE ? "" : s.substring(start, start + len - 1);
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}
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public String minWindow1(String s, String t) {
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HashMap<Character, Integer> need = new HashMap<>();
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for (char c : t.toCharArray()) {
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need.put(c, need.getOrDefault(c, 0) + 1);
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}
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HashMap<Character, Integer> window = new HashMap<>();
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int left = 0;
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int right = 0;
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int count = 0;
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int result = Integer.MAX_VALUE;
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int start = 0;
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while (right < s.length()) {
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char c = s.charAt(right++);
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if (need.containsKey(c)) {
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window.put(c, window.getOrDefault(c, 0) + 1);
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if (window.get(c).equals(need.get(c))) {
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count++;
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}
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}
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while (left < right && count == need.size()) {
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int length = right - left;
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if (length < result) {
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result = length;
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start = left;
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}
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char c1 = s.charAt(left++);
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if (need.containsKey(c1)) {
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Integer num = window.get(c1);
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if (num.equals(need.get(c1))) {
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count--;
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}
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window.put(c1, num - 1);
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}
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}
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}
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return result == Integer.MAX_VALUE ? "" : s.substring(start, start + result);
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}
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public String minWindow2(String s, String t) {
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HashMap<Character, Integer> need = new HashMap<>();
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//遍历t,获取每个字母需要的个数
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for (int i = 0; i < t.length(); i++) {
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char c = t.charAt(i);
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need.put(c, need.getOrDefault(c, 0) + 1);
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}
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HashMap<Character, Integer> window = new HashMap<>();//窗口
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int left = 0;
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int right = 0;
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int diff = need.size();
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int minLength = Integer.MAX_VALUE;
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int start = 0;
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while (right < s.length()) {
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char c = s.charAt(right++);
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if (need.containsKey(c)) {
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Integer count = window.getOrDefault(c, 0);
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if (count.equals(need.get(c) - 1)) {
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diff--;
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}
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window.put(c, count + 1);
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}
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while (diff == 0 && left < right) {
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int curLength = right - left;//由于之前right++了,所以这里不用+1了
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if (curLength < minLength) {
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minLength = curLength;
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start = left;
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}
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char c1 = s.charAt(left++);
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if (need.containsKey(c1)) {
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Integer count = window.getOrDefault(c1, 0);
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if (count.equals(need.get(c1))) {
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diff++;
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}
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window.put(c1, count - 1);
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}
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}
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}
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return minLength == Integer.MAX_VALUE ? "" : s.substring(start, start + minLength);
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}
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//二刷
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public String minWindow3(String s, String t) {
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HashMap<Character, Integer> need = new HashMap<>();
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for (int i = 0; i < t.length(); i++) {
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char c = t.charAt(i);
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need.put(c, need.getOrDefault(c, 0) + 1);
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}
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HashMap<Character, Integer> window = new HashMap<>();
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int left = 0;
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int right = 0;
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int start = 0;
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int len = Integer.MAX_VALUE;
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int diff = need.size();
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while (right < s.length()) {
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char c = s.charAt(right++);
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if (need.containsKey(c)) {
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window.put(c, window.getOrDefault(c, 0) + 1);
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if (window.get(c).equals(need.get(c))) {
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diff--;
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}
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}
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while (diff == 0 && left < right) {
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int cur = right - left;
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if (cur < len) {
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len = cur;
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start = left;
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}
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char c1 = s.charAt(left++);
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if (need.containsKey(c1)) {
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if (window.get(c1).equals(need.get(c1))) {
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diff++;
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}
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window.put(c1, window.get(c1) - 1);
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}
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}
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}
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return len == Integer.MAX_VALUE ? "" : s.substring(start, start + len);
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}
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}
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@ -181,5 +181,53 @@ public class T35_76_MinWindow {
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return minLength == Integer.MAX_VALUE ? "" : s.substring(start, start + minLength);
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}
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//二刷
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public String minWindow3(String s, String t) {
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HashMap<Character, Integer> need = new HashMap<>();
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for (int i = 0; i < t.length(); i++) {
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char c = t.charAt(i);
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need.put(c, need.getOrDefault(c, 0) + 1);
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}
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HashMap<Character, Integer> window = new HashMap<>();
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int left = 0;
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int right = 0;
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int start = 0;
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int len = Integer.MAX_VALUE;
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int diff = need.size();
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while (right < s.length()) {
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char c = s.charAt(right++);
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if (need.containsKey(c)) {
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window.put(c, window.getOrDefault(c, 0) + 1);
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if (window.get(c).equals(need.get(c))) {
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diff--;
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}
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}
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while (diff == 0 && left < right) {
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int cur = right - left;
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if (cur < len) {
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len = cur;
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start = left;
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}
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char c1 = s.charAt(left++);
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if (need.containsKey(c1)) {
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if (window.get(c1).equals(need.get(c1))) {
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diff++;
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}
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window.put(c1, window.get(c1) - 1);
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}
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}
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}
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return len == Integer.MAX_VALUE ? "" : s.substring(start, start + len);
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}
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}
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@ -0,0 +1,33 @@
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package com.markilue.leecode.hot100.second;
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import com.markilue.leecode.tree.TreeNode;
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/**
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*@BelongsProject: Leecode
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*@BelongsPackage: com.markilue.leecode.hot100.second
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*@Author: markilue
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*@CreateTime: 2023-05-08 10:42
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*@Description:
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* TODO 力扣98 验证二叉搜索树:
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* 给你一个二叉树的根节点 root ，判断其是否是一个有效的二叉搜索树。
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* 有效 二叉搜索树定义如下：
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* 节点的左子树只包含 小于 当前节点的数。
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* 节点的右子树只包含 大于 当前节点的数。
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* 所有左子树和右子树自身必须也是二叉搜索树。
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*@Version: 1.0
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*/
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public class T42_98_IsValidBST {
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public boolean isValidBST(TreeNode root) {
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return isValid(root, Long.MIN_VALUE, Long.MAX_VALUE);
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}
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public boolean isValid(TreeNode root, long leftThreshold, long rightThreshold) {
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if (root == null) {
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return true;
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}
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boolean flag = root.val > leftThreshold && root.val < rightThreshold;
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return flag && isValid(root.left, leftThreshold, root.val) && isValid(root.right, root.val, rightThreshold);
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}
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}
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@ -0,0 +1,35 @@
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package com.markilue.leecode.hot100.second;
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import com.markilue.leecode.tree.TreeNode;
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/**
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||||
*@BelongsProject: Leecode
|
||||
*@BelongsPackage: com.markilue.leecode.hot100.second
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*@Author: markilue
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*@CreateTime: 2023-05-08 10:42
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*@Description:
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* TODO 力扣101 对称二叉树:
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* 给你一个二叉树的根节点 root ， 检查它是否轴对称。
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*@Version: 1.0
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*/
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public class T43_101_IsSymmetric {
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public boolean isSymmetric(TreeNode root) {
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if (root == null) {
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return true;
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}
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||||
return isSymmetric(root.left, root.right);
|
||||
}
|
||||
|
||||
public boolean isSymmetric(TreeNode node1, TreeNode node2) {
|
||||
if (node1 == null && node2 == null) {
|
||||
return true;
|
||||
} else if (node1 == null || node2 == null) {
|
||||
return false;
|
||||
} else {
|
||||
return node1.val == node2.val && isSymmetric(node1.left, node2.right) && isSymmetric(node1.right, node2.left);
|
||||
}
|
||||
|
||||
}
|
||||
|
||||
}
|
||||
|
|
@ -0,0 +1,69 @@
|
|||
package com.markilue.leecode.hot100.second;
|
||||
|
||||
import com.markilue.leecode.tree.TreeNode;
|
||||
|
||||
import java.util.ArrayList;
|
||||
import java.util.LinkedList;
|
||||
import java.util.List;
|
||||
|
||||
/**
|
||||
*@BelongsProject: Leecode
|
||||
*@BelongsPackage: com.markilue.leecode.hot100.second
|
||||
*@Author: markilue
|
||||
*@CreateTime: 2023-05-08 11:01
|
||||
*@Description:
|
||||
* TODO 力扣102 二叉树的层序遍历:
|
||||
* 给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。
|
||||
*@Version: 1.0
|
||||
*/
|
||||
public class T44_102_LevelOrder {
|
||||
|
||||
List<List<Integer>> result = new ArrayList<>();
|
||||
|
||||
//dfs递归法
|
||||
public List<List<Integer>> levelOrder(TreeNode root) {
|
||||
levelOrder(root, 0);
|
||||
return result;
|
||||
}
|
||||
|
||||
public void levelOrder(TreeNode node, int level) {
|
||||
if (node == null) return;
|
||||
if (level >= result.size()) {
|
||||
result.add(new ArrayList<>());
|
||||
}
|
||||
result.get(level).add(node.val);
|
||||
levelOrder(node.left, level + 1);
|
||||
levelOrder(node.right, level + 1);
|
||||
}
|
||||
|
||||
|
||||
/**
|
||||
* 队列法
|
||||
* @param root
|
||||
* @return
|
||||
*/
|
||||
public List<List<Integer>> levelOrder1(TreeNode root) {
|
||||
|
||||
|
||||
LinkedList<TreeNode> queue = new LinkedList<>();
|
||||
ArrayList<List<Integer>> result = new ArrayList<>();
|
||||
if (root == null) return result;
|
||||
|
||||
queue.add(root);
|
||||
|
||||
while (!queue.isEmpty()) {
|
||||
int size = queue.size();
|
||||
ArrayList<Integer> cur = new ArrayList<>();
|
||||
for (int i = 0; i < size; i++) {
|
||||
TreeNode treeNode = queue.pollFirst();
|
||||
cur.add(treeNode.val);
|
||||
if (treeNode.left != null) queue.addLast(treeNode.left);
|
||||
if (treeNode.right != null) queue.addLast(treeNode.right);
|
||||
}
|
||||
result.add(cur);
|
||||
}
|
||||
|
||||
return result;
|
||||
|
||||
}
|
||||
}
|
||||
|
|
@ -0,0 +1,34 @@
|
|||
package com.markilue.leecode.hot100.second;
|
||||
|
||||
import com.markilue.leecode.tree.TreeNode;
|
||||
|
||||
/**
|
||||
*@BelongsProject: Leecode
|
||||
*@BelongsPackage: com.markilue.leecode.hot100.second
|
||||
*@Author: markilue
|
||||
*@CreateTime: 2023-05-08 11:15
|
||||
*@Description:
|
||||
* TODO 力扣104 二叉树的最大深度:
|
||||
* 给定一个二叉树，找出其最大深度。
|
||||
* 二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。
|
||||
* 说明: 叶子节点是指没有子节点的节点。
|
||||
*@Version: 1.0
|
||||
*/
|
||||
public class T45_104_MaxDepth {
|
||||
|
||||
int max = 0;
|
||||
|
||||
public int maxDepth(TreeNode root) {
|
||||
maxDepth(root, 1);
|
||||
return max;
|
||||
}
|
||||
|
||||
public void maxDepth(TreeNode root, int depth) {
|
||||
if (root == null) {
|
||||
return;
|
||||
}
|
||||
if (max < depth) max = depth;
|
||||
maxDepth(root.left, depth + 1);
|
||||
maxDepth(root.right, depth + 1);
|
||||
}
|
||||
}
|
||||
|
|
@ -1,17 +1,28 @@
|
|||
package com.markilue.leecode.hot100.second;
|
||||
|
||||
import com.markilue.leecode.tree.TreeNode;
|
||||
import com.markilue.leecode.tree.TreeUtils;
|
||||
import org.junit.Test;
|
||||
|
||||
/**
|
||||
* @BelongsProject: Leecode
|
||||
* @BelongsPackage: com.markilue.leecode.hot100.second
|
||||
* @Author: marklue
|
||||
* @CreateTime: 2023/4/16 11:15
|
||||
* @Description: TODO 力扣105 根据前序遍历和中序遍历构造树
|
||||
* @Description:
|
||||
* TODO 力扣105 根据前序遍历和中序遍历构造树
|
||||
* 给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。
|
||||
* @Version: 1.0
|
||||
*/
|
||||
public class T46_105_BuildTree {
|
||||
|
||||
@Test
|
||||
public void test() {
|
||||
int[] preorder = {3, 9, 20, 15, 7};
|
||||
int[] inorder = {9, 3, 15, 20, 7};
|
||||
TreeUtils.printTreeByLevel(buildTree1(preorder,inorder));
|
||||
}
|
||||
|
||||
int index = 0;
|
||||
|
||||
public TreeNode buildTree(int[] preorder, int[] inorder) {
|
||||
|
|
@ -53,4 +64,55 @@ public class T46_105_BuildTree {
|
|||
return -1;//没找到
|
||||
|
||||
}
|
||||
|
||||
|
||||
public TreeNode buildTree1(int[] preorder, int[] inorder) {
|
||||
return buildSubTree(preorder, inorder, 0, inorder.length);
|
||||
}
|
||||
|
||||
public TreeNode buildSubTree(int[] preorder, int[] inorder, int inStart, int inEnd) {
|
||||
if (inStart >= inEnd) return null;
|
||||
TreeNode root = new TreeNode(preorder[index]);
|
||||
int index1 = find(inorder, inStart, inEnd, preorder[index++]);
|
||||
root.left = buildSubTree(preorder, inorder, inStart, index1);//注意这里是index1而不是index1减1，因为这里是边界的意思，和上面inorder.length对应
|
||||
root.right = buildSubTree(preorder, inorder, index1 + 1, inEnd);
|
||||
return root;
|
||||
}
|
||||
|
||||
public int find(int[] inorder, int start, int end, int num) {
|
||||
|
||||
for (int i = start; i < end; i++) {
|
||||
if (inorder[i] == num) {
|
||||
return i;
|
||||
}
|
||||
}
|
||||
|
||||
return -1;
|
||||
}
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
//官方最快
|
||||
private int in = 0;
|
||||
private int pre = 0;
|
||||
|
||||
public TreeNode buildTree3(int[] preorder, int[] inorder) {
|
||||
return build(preorder, inorder, Integer.MIN_VALUE);
|
||||
}
|
||||
|
||||
private TreeNode build(int[] preorder, int[] inorder, int stop) {
|
||||
if (pre >= preorder.length)
|
||||
return null;
|
||||
if (inorder[in] == stop) {//合理使用stop即到了stop就到了右边界
|
||||
in++;
|
||||
return null;
|
||||
}
|
||||
|
||||
TreeNode node = new TreeNode(preorder[pre++]);
|
||||
node.left = build(preorder, inorder, node.val);
|
||||
node.right = build(preorder, inorder, stop);
|
||||
return node;
|
||||
}
|
||||
}
|
||||
|
|
|
|||
Loading…
Reference in New Issue