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leecode更新

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markilue 2023-02-28 14:30:04 +08:00
parent 667ca9a0b9
commit 381f8efdde
4 changed files with 369 additions and 3 deletions
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2
Big_data_example/Flink/src/main/java/day05/Example4.java
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@ -60,7 +60,7 @@ public class Example4 {
@Override @Override
public void process(String s, Context context, Iterable<Tuple2<String, Long>> iterable, Collector<String> collector) throws Exception { public void process(String s, Context context, Iterable<Tuple2<String, Long>> iterable, Collector<String> collector) throws Exception {
//初始化一个窗口状态变量，注意:窗口状态变量的可见范围是当前窗口 //初始化一个窗口状态变量，注意:窗口状态变量的可见范围是当前窗口(无论窗口是否被销毁)
ValueState<Boolean> firstCalculate = context.windowState().getState(new ValueStateDescriptor<Boolean>("first", Types.BOOLEAN)); ValueState<Boolean> firstCalculate = context.windowState().getState(new ValueStateDescriptor<Boolean>("first", Types.BOOLEAN));
if (firstCalculate.value() == null) { if (firstCalculate.value() == null) {

2
Big_data_example/Flink/src/main/java/day08/Example9.java
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@ -83,8 +83,6 @@ public class Example9 {
+I[Bob, 2, 1970-01-01T15:00] +I[Bob, 2, 1970-01-01T15:00]
+I[Mary, 1, 1970-01-01T15:00] +I[Mary, 1, 1970-01-01T15:00]
+I[liz, 1, 1970-01-01T15:00] +I[liz, 1, 1970-01-01T15:00]
*/ */

191
Leecode/src/main/java/com/markilue/leecode/hot100/T05_LongestPalindrome.java Normal file
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@ -0,0 +1,191 @@
package com.markilue.leecode.hot100;
import org.junit.Test;
/**
*@BelongsProject: Leecode
*@BelongsPackage: com.markilue.leecode.hot100
*@Author: markilue
*@CreateTime: 2023-02-28 10:10
*@Description:
* TODO 力扣5题 最长回文子串:
* 给你一个字符串 s，找到 s 中最长的回文子串。
* 如果字符串的反序与原始字符串相同，则该字符串称为回文字符串。
*@Version: 1.0
*/
public class T05_LongestPalindrome {
@Test
public void test() {
String s = "babad";
System.out.println(longestPalindrome1(s));
}
/**
* 思路:最长回文子串，则不能删除，一旦不是回文子串则直接归零
* TODO DP五部曲:
* 1.dp定义:dp[i][j]表示以j开始以i结尾的子串是不是回文子串
* 2.dp状态转移方程:
* 1.char[i]==char[j]
* dp[i][j]=dp[i-1][j+1]
* 3.dp初始化: dp[i][i]=true
* 4.dp遍历顺序:从上往下
* 5.dp举例推导:
* 速度击败51.19% 内存击败7.46% 79ms
* 修改后 62ms
* @param s
* @return
*/
public String longestPalindrome(String s) {
char[] chars = s.toCharArray();
boolean[][] dp = new boolean[chars.length][chars.length];
int maxLength = 0;
int begin = 0;
int end = 0;
for (int i = 0; i < chars.length; i++) {
for (int j = 0; j <= i; j++) {
if (chars[i] == chars[j]) {
if (i <= j + 1) {
dp[i][j] = true;
} else {
dp[i][j] = dp[i - 1][j + 1];
}
}
if (dp[i][j]) {
if (maxLength < i - j + 1) {
maxLength = i - j + 1;
begin = j;
end = i + 1;
}
}
}
}
return s.substring(begin, end);
}
public boolean isPalindrome(char[] chars, int start, int end) {
while (start < end) {
if (chars[start++] != chars[end--]) return false;
}
return true;
}
/**
* 一维dp优化
* 速度击败54.74% 内存击败77.79% 42ms
* @param s
* @return
*/
public String longestPalindrome1(String s) {
char[] chars = s.toCharArray();
boolean[] dp = new boolean[chars.length];
int maxLength = 0;
int begin = 0;
for (int i = 0; i < chars.length; i++) {
for (int j = 0; j <= i; j++) {
if (chars[i] == chars[j]) {
if (i <= j + 1) {
dp[j] = true;
} else {
dp[j] = dp[j + 1];
}
}else {
dp[j]=false;
}
if (dp[j]&&maxLength < i - j + 1) {
maxLength = i - j + 1;
begin = j;
}
}
}
return s.substring(begin, begin+maxLength);
}
/**
* 官方最快:
* 分别计算考虑左边；考虑右边；和以i为中心，本质上应该还是O(N^2)
* 速度击败93.2% 内存击败78.66% 7ms
* @param s
* @return
*/
public String longestPalindrome2(String s) {
char[] charStr = s.toCharArray();
int maxlen = 0;
int maxStart = 0;
for (int i = 0; i < charStr.length; i++) {
int len = 1;
int left = i - 1;
int right = i + 1;
while (left >= 0 && charStr[i] == charStr[left]) {//查找左边
len++;
left--;
}
while (right < charStr.length && charStr[i] == charStr[right]) {//查找右边
len++;
right++;
}
while (left >= 0 && right < charStr.length && charStr[left] == charStr[right]) {//以i为中心
len += 2;
left--;
right++;
}
if (len > maxlen) {
maxlen = len;
maxStart = left;
}
}
return s.substring(maxStart + 1, maxlen + maxStart + 1);
}
/**
* 中心扩展法:看以两个为中心，和以一个为中心是不是回文子串
* @param s
* @return
*/
public String longestPalindrome3(String s) {
if (s == null || s.length() < 1) {
return "";
}
int start = 0, end = 0;
for (int i = 0; i < s.length(); i++) {
int len1 = expandAroundCenter(s, i, i);
int len2 = expandAroundCenter(s, i, i + 1);
int len = Math.max(len1, len2);
if (len > end - start) {
start = i - (len - 1) / 2;
end = i + len / 2;
}
}
return s.substring(start, end + 1);
}
public int expandAroundCenter(String s, int left, int right) {
while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
--left;
++right;
}
return right - left - 1;
}
}

177
Leecode/src/main/java/com/markilue/leecode/hot100/T06_Convert.java Normal file
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@ -0,0 +1,177 @@
package com.markilue.leecode.hot100;
import com.sun.java.swing.plaf.windows.WindowsDesktopIconUI;
import org.junit.Test;
import java.util.Arrays;
/**
*@BelongsProject: Leecode
*@BelongsPackage: com.markilue.leecode.hot100
*@Author: markilue
*@CreateTime: 2023-02-28 11:52
*@Description:
* TODO 力扣6题 N字形变形:
* 将一个给定字符串 s 根据给定的行数 numRows ，以从上往下、从左到右进行 Z 字形排列。
* 比如输入字符串为 "PAYPALISHIRING" 行数为 3 时，排列如下：
* P A H N
* A P L S I I G
* Y I R
* 之后，你的输出需要从左往右逐行读取，产生出一个新的字符串，比如："PAHNAPLSIIGYIR"。
* 请你实现这个将字符串进行指定行数变换的函数：
* string convert(string s, int numRows);
*@Version: 1.0
*/
public class T06_Convert {
@Test
public void test() {
// String s = "ABC";
String s = "PAYPALISHIRING";
System.out.println(convert3(s, 3));
}
/**
* 思路:先计算一共需要多少列才能记录
* 速度击败21.1% 内存击败22.88% 26ms
* @param s
* @param numRows
* @return
*/
public String convert(String s, int numRows) {
//先计算一共需要多少列才能记录
int length = s.length();
if (numRows >= length || numRows == 1) return s;
char[] chars1 = s.toCharArray();
int index = 0;
int col = length / 2;
char[][] chars = new char[numRows][col + 1];
for (int k = 0; k < length / (numRows - 1) && index < length; k++) {
int i = 0;
int j = 0;
for (; i < numRows - 1 && index < length; i++) {
chars[i][k * (numRows - 1)] = chars1[index++];
}
for (; i > 0 && index < length; i--, j++) {
chars[i][k * (numRows - 1) + j] = chars1[index++];
}
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < chars.length; i++) {
for (int j = 0; j < chars[i].length; j++) {
if (chars[i][j] != '\u0000')
sb.append(chars[i][j]);
}
}
return sb.toString();
}
/**
* 官方思路:本质上和自己的是一致的，但是判断依据进行了简化
* 速度击败18.75% 内存击败12.36% 35ms
* @param s
* @param numRows
* @return
*/
public String convert1(String s, int numRows) {
int n = s.length(), r = numRows;
if (r == 1 || r >= n) {
return s;
}
int t = r * 2 - 2;
int c = (n + t - 1) / t * (r - 1);
char[][] mat = new char[r][c];
for (int i = 0, x = 0, y = 0; i < n; ++i) {
mat[x][y] = s.charAt(i);
if (i % t < r - 1) {
++x; // 向下移动
} else {
--x;
++y; // 向右上移动
}
}
StringBuffer ans = new StringBuffer();
for (char[] row : mat) {
for (char ch : row) {
if (ch != 0) {
ans.append(ch);
}
}
}
return ans.toString();
}
/**
* 思路:直接使用stringbuilder进行append
* 速度击败43.99% 内存击败56.51% 8ms
* @param s
* @param numRows
* @return
*/
public String convert2(String s, int numRows) {
int n = s.length(), r = numRows;
if (r == 1 || r >= n) {
return s;
}
StringBuffer[] mat = new StringBuffer[r];
for (int i = 0; i < r; ++i) {
mat[i] = new StringBuffer();
}
for (int i = 0, x = 0, t = r * 2 - 2; i < n; ++i) {
mat[x].append(s.charAt(i));
if (i % t < r - 1) {
++x;
} else {
--x;
}
}
StringBuffer ans = new StringBuffer();
for (StringBuffer row : mat) {
ans.append(row);
}
return ans.toString();
}
/**
* 官方直接构造法:
* 把其中的0,1,2,3改成循环的i变量，就是代码。 吐槽一句：这个表格编辑的时候看起来整齐，预览却不整齐，调整了好半天，官方大大能不能优化一下。
* 0 0+t 0+2t 0+3t
* 1 t-1 1+t 0+2t-1 1+2t 0+3t-1 1+3t
* 2 t-2 2+t 0+2t-2 2+2t 0+3t-2 2+3t
* 3 3+t 3+2t 3+3t
* 好像暂时有错误
* @param s
* @param numRows
* @return
*/
public String convert3(String s, int numRows) {
int n = s.length(), r = numRows;
if (r == 1 || r >= n) {
return s;
}
StringBuilder sb = new StringBuilder();
int t = r * 2 - 2;
for (int i = 0; i < r; i++) {// 枚举矩阵的行
for (int j = 0; j + i < n; j += t) {// 枚举每个周期的起始下标
sb.append(s.charAt(j + i));// 当前周期的第一个字符
if (0 < i && i < r - 1 && j + t - i < n) {
sb.append(s.charAt(j + t - 1));// 当前周期的第二个字符
}
}
}
return sb.toString();
}
}