leecode更新
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markilue
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package com.markilue.leecode.hot100.second;
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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.second
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*@Author: markilue
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*@CreateTime: 2023-04-17 09:50
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*@Description:
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* TODO 力扣5题 最长回文子串:
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* 给你一个字符串 s，找到 s 中最长的回文子串。
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* 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。
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*@Version: 1.0
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*/
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public class T05_5_LongestPalindrome {
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@Test
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public void test() {
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String s = "cbbd";
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System.out.println(longestPalindrome(s));
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}
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/**
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* 思路:动态规划判断一个子串是不是回文，如果是判断他是否更长
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* @param s
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* @return
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*/
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public String longestPalindrome(String s) {
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boolean[][] dp = new boolean[s.length() + 1][s.length() + 1];//<start,end>
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dp[0][0] = true;
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char[] chars = s.toCharArray();
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int start = 0;
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int length = 0;
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for (int i = 1; i < dp.length; i++) {//end
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for (int j = i; j >= 1; j--) {//start
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if (chars[i - 1] == chars[j - 1]) {
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if (i <= j + 1) {
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dp[i][j] = true;
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} else {
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dp[i][j] = dp[i - 1][j + 1];
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}
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}
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if (dp[i][j] && i - j + 1 > length) {
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start = j-1;
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length = i - j + 1;
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}
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}
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}
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return s.substring(start, start + length);
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}
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}
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package com.markilue.leecode.hot100.second;
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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: 2023-04-17 10:09
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*@Description:
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* TODO 力扣11 盛最多的水的容器:
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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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*@Version: 1.0
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*/
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public class T09_11_MaxArea {
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/**
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* 思路:双指针法——哪边矮走哪一边
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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 left = 0;
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int right = height.length - 1;
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int maxArea = 0;
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while (left != right) {
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int min = Math.min(height[left], height[right]);
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maxArea = Math.max(maxArea, min * (right - left));
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if (height[left] < height[right]) {
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left++;
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} else {
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right--;
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}
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}
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return maxArea;
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}
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/**
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* 思路:双指针法——更快
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* @param height
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* @return
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*/
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public int maxArea1(int[] height) {
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int left = 0;
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int right = height.length - 1;
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int maxArea = 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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maxArea = Math.max(maxArea, 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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maxArea = Math.max(maxArea, 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 maxArea;
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}
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}
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package com.markilue.leecode.hot100.second;
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import org.junit.Test;
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import java.util.LinkedList;
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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: 2023-04-17 11:35
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*@Description:
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* TODO 力扣32 最长有效括号:
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* 给你一个只包含 '(' 和 ')' 的字符串，找出最长有效（格式正确且连续）括号子串的长度。
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*@Version: 1.0
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*/
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public class T19_32_LongestValidParentheses {
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@Test
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public void test(){
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String s = ")()())";
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System.out.println(longestValidParentheses(s));
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}
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/**
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* 使用stack记录对应入栈的索引
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* @param s
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* @return
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*/
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public int longestValidParentheses(String s) {
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LinkedList<Integer> stack = new LinkedList<>();
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int length = 0;
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stack.push(-1);
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char[] chars = s.toCharArray();
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for (int i = 0; i < chars.length; i++) {
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if (chars[i] == '(') {
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stack.push(i);
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} else {
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stack.pop();
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if (stack.isEmpty()) {
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stack.push(i);
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}else {
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length = Math.max(i-stack.peek() , length);
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}
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}
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}
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return length;
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}
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}
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package com.markilue.leecode.hot100.second;
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import com.sun.media.sound.RIFFInvalidDataException;
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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.second
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*@Author: markilue
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*@CreateTime: 2023-04-17 11:15
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*@Description:
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* TODO 力扣42 接雨水:
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* 给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。
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*@Version: 1.0
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*/
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public class T23_42_Trap {
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@Test
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public void test(){
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int[] height={0,1,0,2,1,0,1,3,2,1,2,1};
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System.out.println(trap(height));
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}
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/**
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* 双指针法:谁低走谁
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* @param height
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* @return
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*/
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public int trap(int[] height) {
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int leftMax = height[0];
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int rightMax = height[height.length - 1];
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int sum = 0;
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int left = 0;
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int right = height.length - 1;
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while (left < right) {
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leftMax = Math.max(leftMax, height[left]);
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rightMax = Math.max(rightMax, height[right]);
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if (height[left] < height[right]) {
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//左边一定的积水
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sum += Math.min(leftMax, rightMax) - height[left];
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left++;
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} else {
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sum += Math.min(leftMax, rightMax) - height[right];
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right--;
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}
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}
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return sum;
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}
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}
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