leecode更新
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markilue
parent
381f8efdde
commit
ee7a15d432
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@ -33,7 +33,7 @@ public class Example7 {
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env.setParallelism(1);
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env
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.readTextFile("D:\\example\\self_example\\Big_data_example\\Flink\\src\\main\\resources\\UserBehavior.csv")
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.readTextFile("E:\\self_example\\Big_data_example\\Flink\\src\\main\\resources\\UserBehavior.csv")
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.map(new MapFunction<String, UserBehavior>() {
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@Override
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public UserBehavior map(String value) throws Exception {
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@ -0,0 +1,92 @@
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package com.markilue.leecode.hot100;
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import com.sun.deploy.util.StringUtils;
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import org.junit.Test;
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/**
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*@BelongsProject: Leecode
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*@BelongsPackage: com.markilue.leecode.hot100
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*@Author: markilue
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*@CreateTime: 2023-03-01 10:11
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*@Description:
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* TODO 力扣7题 整数反转:
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* 给你一个 32 位的有符号整数 x ，返回将 x 中的数字部分反转后的结果。
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* 如果反转后整数超过 32 位的有符号整数的范围 [−231, 231 − 1] ，就返回 0。
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* 假设环境不允许存储 64 位整数（有符号或无符号）。
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*@Version: 1.0
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*/
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public class T07_Reverse {
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@Test
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public void test() {
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// String s = "00205";
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// System.out.println(Integer.parseInt(s));205
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// int x=-120;
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int x = -2147483648;
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System.out.println(reverse1(x));
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}
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/**
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* 思路:把它转化为字符串，在进行反转
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* 速度击败17.3% 内存击败53.2% 2ms
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* @param x
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* @return
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*/
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public int reverse(int x) {
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boolean flag = true;//记录这个数是否是正数
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if (x < 0) flag = false;
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String s = "" + Math.abs((long) x);//这里必须要转为long,不然Integer.Min不能进行反转
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char[] chars = s.toCharArray();
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//反转字符串
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reverseString(chars);
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long result = Long.parseLong(new String(chars));
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if (!flag) result = -result;
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if (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE) return 0;
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return (int) result;
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}
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public void reverseString(char[] chars) {
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int left = 0;
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int right = chars.length - 1;
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char temp;
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while (left < right) {
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temp = chars[left];
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chars[left] = chars[right];
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chars[right] = temp;
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left++;
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right--;
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}
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}
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/**
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* 思路:考虑直接把它当整数进行反转
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* 速度击败100% 内存击败56.5% 0ms
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* @param x
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* @return
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*/
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public int reverse1(int x) {
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boolean flag = true;//记录这个数是否是正数
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if (x < 0) flag = false;
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long y = (long) x;
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long result = 0;
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while (y != 0) {
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result = result * 10 + y % 10;
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y /= 10;
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}
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if (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE) return 0;
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return (int) result;
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}
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}
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@ -0,0 +1,151 @@
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package com.markilue.leecode.hot100;
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import org.junit.Test;
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import java.util.HashMap;
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import java.util.Map;
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/**
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*@BelongsProject: Leecode
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*@BelongsPackage: com.markilue.leecode.hot100
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*@Author: markilue
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*@CreateTime: 2023-03-01 10:53
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*@Description:
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* TODO 力扣8题 字符串转换整数(atoi):
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* 请你来实现一个 myAtoi(string s) 函数，使其能将字符串转换成一个 32 位有符号整数（类似 C/C++ 中的 atoi 函数）。
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* 函数 myAtoi(string s) 的算法如下：
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* 读入字符串并丢弃无用的前导空格
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* 检查下一个字符（假设还未到字符末尾）为正还是负号，读取该字符（如果有）。 确定最终结果是负数还是正数。 如果两者都不存在，则假定结果为正。
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* 读入下一个字符，直到到达下一个非数字字符或到达输入的结尾。字符串的其余部分将被忽略。
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* 将前面步骤读入的这些数字转换为整数（即，"123" -> 123， "0032" -> 32）。如果没有读入数字，则整数为 0 。必要时更改符号（从步骤 2 开始）。
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* 如果整数数超过 32 位有符号整数范围 [−231, 231 − 1] ，需要截断这个整数，使其保持在这个范围内。具体来说，小于 −231 的整数应该被固定为 −231 ，大于 231 − 1 的整数应该被固定为 231 − 1 。
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* 返回整数作为最终结果。
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* 注意：
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* 本题中的空白字符只包括空格字符 ' ' 。
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* 除前导空格或数字后的其余字符串外，请勿忽略 任何其他字符。
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*@Version: 1.0
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*/
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public class T08_MyAtoi {
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@Test
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public void test() {
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String s = " ss-125-sa7894e ";
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String s1 = " words and 987 ";
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String s2 = " wor98ds and 7 ";
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String s3 = " 9words and 87 ";
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String s5 = " 9 8words and 87 ";
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String s6 = " -9 8words and 87 ";
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String s4 = "2147483648";
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String s7 = "20000000000000000000";
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// System.out.println(myAtoi(s));
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// System.out.println(myAtoi(s1));
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// System.out.println(myAtoi(s2));
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// System.out.println(myAtoi(s3));
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System.out.println(myAtoi(s4));
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System.out.println(myAtoi(s5));
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System.out.println(myAtoi(s6));
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System.out.println(myAtoi(s7));
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}
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/**
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* 思路:利用StringBuilder记录
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* 速度击败100% 内存击败84.23%
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* @param s
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* @return
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*/
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public int myAtoi(String s) {
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char[] chars = s.toCharArray();
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StringBuilder sb = new StringBuilder();
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boolean flag = false;//记录是否已经开始匹配到数字 :避免后续有 ‘-’的情况
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for (char aChar : chars) {
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if (((aChar == '-' || aChar == '+') && !flag) || (aChar - '0' >= 0 && aChar - '0' <= 9)) {
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//是数字；或者是开头的-
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sb.append(aChar);
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flag = true;
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} else if (!flag && aChar == ' ') {
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continue;
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} else {
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break;
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}
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}
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if (sb.length() == 0 || (sb.length() == 1 && (sb.charAt(0) == '-' || sb.charAt(0) == '+'))) return 0;
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// long result = Long.parseLong(sb.toString());//超出长度
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boolean flag1 = true;//记录这个数是不是正数
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long result = 0;
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int startIndex = 0;
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if (sb.charAt(0) == '-') {
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flag1 = false;
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startIndex = 1;
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} else if (sb.charAt(0) == '+') {
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startIndex = 1;
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}
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for (int i = startIndex; i < sb.length(); i++) {
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if (flag1) {
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result = result * 10 + (sb.charAt(i) - '0');
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} else {
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result = result * 10 - (sb.charAt(i) - '0');
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}
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if (result < Integer.MIN_VALUE) return Integer.MIN_VALUE;
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if (result > Integer.MAX_VALUE) return Integer.MAX_VALUE;
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}
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return (int) result;
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}
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/**
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* 官方解法:有限状态机
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* @param str
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* @return
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*/
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public int myAtoi1(String str) {
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Automaton automaton = new Automaton();
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int length = str.length();
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for (int i = 0; i < length; ++i) {
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automaton.get(str.charAt(i));
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}
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return (int) (automaton.sign * automaton.ans);
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}
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class Automaton {
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public int sign = 1;
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public long ans = 0;
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private String state = "start";
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private Map<String, String[]> table = new HashMap<String, String[]>() {{
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put("start", new String[]{"start", "signed", "in_number", "end"});
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put("signed", new String[]{"end", "end", "in_number", "end"});
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put("in_number", new String[]{"end", "end", "in_number", "end"});
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put("end", new String[]{"end", "end", "end", "end"});
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}};
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public void get(char c) {
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state = table.get(state)[get_col(c)];
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if ("in_number".equals(state)) {
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ans = ans * 10 + c - '0';
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ans = sign == 1 ? Math.min(ans, (long) Integer.MAX_VALUE) : Math.min(ans, -(long) Integer.MIN_VALUE);
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} else if ("signed".equals(state)) {
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sign = c == '+' ? 1 : -1;
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}
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}
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private int get_col(char c) {
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if (c == ' ') {
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return 0;
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}
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if (c == '+' || c == '-') {
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return 1;
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}
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if (Character.isDigit(c)) {
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return 2;
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}
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return 3;
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}
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}
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}
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@ -0,0 +1,163 @@
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package com.markilue.leecode.hot100;
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import org.junit.Test;
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/**
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*@BelongsProject: Leecode
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*@BelongsPackage: com.markilue.leecode.hot100
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*@Author: markilue
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*@CreateTime: 2023-03-01 12:09
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*@Description:
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* TODO 力扣11题 盛最多的水:
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* 给定一个长度为 n 的整数数组 height 。有 n 条垂线，第 i 条线的两个端点是 (i, 0) 和 (i, height[i]) 。
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* 找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。
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* 返回容器可以储存的最大水量。
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* 说明：你不能倾斜容器。
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*@Version: 1.0
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*/
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public class T09_MaxArea {
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@Test
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public void test(){
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int[] height={1,8,6,2,5,4,8,3,7};
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int[] height1={1,0,0,0,0,0,0,2,2};
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System.out.println(maxArea1(height));
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}
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/**
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* 思路:似乎可以使用动态规划:
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* TODO DP五部曲:
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* 1.dp定义: dp[i][j]表示以i开头以j结尾的容器能盛多少水
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* 2.dp状态转移方程:
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* 1.dp[i][j]<min(nums[i],nums[j-1])
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* dp[i][j]=dp[i][j-1]-(j-1-i)*(min-nums[j])+nums[j]
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* 2.min(nums[i],nums[j-1])<dp[i][j]<max(nums[i],nums[j-1])&& min=nums[j-1]
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* dp[i][j]=dp[i][j-1]+(j-1-i)*(nums[j]-min)+nums[j]
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* 3. min=nums[i]
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* dp[i][j]=dp[i][j-1]+nums[i]
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* 4.dp[i][j]>max(nums[i],nums[j-1]) && min=nums[j-1]
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* dp[i][j]=dp[i][j-1]+(j-1-i)*(max-min)+max
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* 3.dp初始化:dp[i][i]=0; dp[i][i+1]=min(nums[i],nums[i+1])
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* 4.dp遍历顺序:
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* 5.dp举例推导:
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* [1 8 6 2 5 4 8 3 7]
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* 1 0 1 2
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* 8
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* 6
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* 2
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* 5
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* 4
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* 8
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* 3
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* 7
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* @param height
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* @return
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*/
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public int maxArea(int[] height) {
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int[][] dp = new int[height.length][height.length];
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// for (int i = 0; i < dp.length; i++) {
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// dp[i][i] = 0;
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// }
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int result = 0;
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for (int i = 0; i < dp.length; i++) {
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for (int j = i + 1; j < dp[0].length; j++) {
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int min = Math.min(height[i], height[j - 1]);
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int max = Math.max(height[i], height[j - 1]);
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if (height[j] < Math.min(height[i], height[j - 1])) {
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dp[i][j] = dp[i][j - 1] - (j - 1 - i) * (min - height[j]) + height[j];
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} else if (min == height[i]) {
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dp[i][j] = dp[i][j - 1] + height[i];
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} else if (min <= height[j] && height[j] <= max) {
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dp[i][j] = dp[i][j - 1] + (j - 1 - i) * (height[j] - min) + height[j];
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} else {
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dp[i][j] = dp[i][j - 1] + (j - 1 - i) * (max - min) + max;
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}
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if(result<dp[i][j]) result=dp[i][j];
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}
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}
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return result;
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}
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public int maxArea1(int[] height) {
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int[] dp=new int[height.length];
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// for (int i = 0; i < dp.length; i++) {
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// dp[i][i] = 0;
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// }
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int result = 0;
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for (int i = 0; i < dp.length; i++) {
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dp=new int[height.length];
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for (int j = i + 1; j < dp.length; j++) {
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int min = Math.min(height[i], height[j - 1]);
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int max = Math.max(height[i], height[j - 1]);
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if (height[j] < Math.min(height[i], height[j - 1])) {
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dp[j] = dp[j - 1] - (j - 1 - i) * (min - height[j]) + height[j];
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} else if (min == height[i]) {
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dp[j] = dp[j - 1] + height[i];
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} else if (min <= height[j] && height[j] <= max) {
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dp[j] = dp[j - 1] + (j - 1 - i) * (height[j] - min) + height[j];
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} else {
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dp[j] = dp[j - 1] + (j - 1 - i) * (max - min) + max;
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}
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if (result < dp[j]) result = dp[j];
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}
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}
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return result;
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}
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/**
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* 官方双指针法，本质上还是让分别上让左右两个为界，计算雨水，得出最大值
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* 速度击败59.94% 内存击败56.49% 4ms
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* @param height
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* @return
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*/
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public int maxArea2(int[] height) {
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int l = 0, r = height.length - 1;
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int ans = 0;
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while (l < r) {
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int area = Math.min(height[l], height[r]) * (r - l);
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ans = Math.max(ans, area);
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if (height[l] <= height[r]) {
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++l;
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}
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else {
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--r;
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}
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}
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return ans;
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}
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/**
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* 官方最快:
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* 本质上就是在官方题解的方法上，增加了一层while循环，减少判断次数
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* 速度击败100% 内存击败29.31% 1ms
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* @param height
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* @return
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*/
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public int maxArea3(int[] height) {
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if(height == null || height.length == 0) return 0;
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int left = 0, right = height.length - 1,area = 0;
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while(left < right){
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if(height[left] <= height[right]){
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int pre = height[left];
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area = Math.max(area, pre * (right - left));
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while(left < right && height[left] <= pre) left++;
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}else{
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int pre = height[right];
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area = Math.max(area, pre * (right - left));
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while(left < right && height[right] <= pre) right--;
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}
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}
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return area;
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}
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}
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