leecode更新
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markilue
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33c16119c6
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0785f33422
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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: 2023-04-26 14:21
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*@Description: TODO 力扣70 爬楼梯:
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*@Version: 1.0
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*/
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public class T32_70_ClimbStairs {
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public int climbStairs(int n) {
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int[] dp = new int[n + 1];
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dp[0] = 1;
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dp[1] = 1;
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for (int i = 2; i < dp.length; i++) {
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dp[i] = dp[i - 1] + dp[i - 2];//走一步 或 走两步
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}
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return dp[n];
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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.Arrays;
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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: 2023-04-26 15:10
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*@Description:
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* TODO 力扣72 编辑距离:
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* 给你两个单词 word1 和 word2， 请返回将 word1 转换成 word2 所使用的最少操作数 。
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* 你可以对一个单词进行如下三种操作：
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* 插入一个字符
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* 删除一个字符
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* 替换一个字符
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*@Version: 1.0
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*/
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public class T33_72_MinDistance {
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@Test
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public void test() {
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String word1 = "horse";
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String word2 = "ros";
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System.out.println(minDistance1(word1, word2));
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}
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/**
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* 动态规划法:
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* DP[i][j]表示使得word1[0-i]和word2[0-j]相同的最小操作数
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* @param word1
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* @param word2
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* @return
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*/
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public int minDistance(String word1, String word2) {
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int[][] dp = new int[word1.length() + 1][word2.length() + 1];
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for (int i = 1; i < dp[0].length; i++) {
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dp[0][i] = i;
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}
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for (int i = 1; i < dp.length; i++) {
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dp[i][0] = i;
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}
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for (int i = 1; i < dp.length; i++) {
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for (int j = 1; j < dp[0].length; j++) {
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if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
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dp[i][j] = dp[i - 1][j - 1];
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} else {
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dp[i][j] = Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;//删除，插入，替换
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}
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}
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}
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return dp[word1.length()][word2.length()];
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}
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int[][] memo;
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/**
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* 回溯+记忆化搜索:
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* 速度击败100% 内存击败76.8% 2ms
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* @param word1
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* @param word2
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* @return
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*/
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public int minDistance1(String word1, String word2) {
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memo = new int[word1.length()][word2.length()];
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for (int i = 0; i < memo.length; i++) {
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Arrays.fill(memo[i], -1);
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}
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return dp(word1, word2, word1.length()-1, word2.length()-1);
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}
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private int dp(String word1, String word2, int i, int j) {
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if (i == -1) {
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return j+1;
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}
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if (j == -1) {
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return i+1;
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}
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if (memo[i][j] != -1) {
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return memo[i][j];
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}
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if (word1.charAt(i) == word2.charAt(j)) {
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memo[i][j] = dp(word1, word2, i - 1, j - 1);
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} else {
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memo[i][j] = Math.min(Math.min(dp(word1, word2, i - 1, j - 1), dp(word1, word2, i - 1, j)), dp(word1, word2, i, j - 1))+1;
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}
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return memo[i][j];
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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.Arrays;
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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: 2023-04-27 10:25
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*@Description:
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* TODO 力扣75 颜色分类:
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* 给定一个包含红色、白色和蓝色、共 n 个元素的数组 nums ，原地对它们进行排序，使得相同颜色的元素相邻，并按照红色、白色、蓝色顺序排列。
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* 我们使用整数 0、 1 和 2 分别表示红色、白色和蓝色。
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* 必须在不使用库内置的 sort 函数的情况下解决这个问题。
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*@Version: 1.0
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*/
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public class T34_75_SortColors {
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@Test
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public void test() {
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int[] nums = {2,0,2,1,1,0};
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sortColors(nums);
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System.out.println(Arrays.toString(nums));
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}
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//遇上0就交换，遇上1就交换1
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public void sortColors(int[] nums) {
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int index0 = 0;
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int index1 = 0;
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for (int i = 0; i < nums.length; i++) {
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if (nums[i] == 0) {
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swap(nums, i, index0);
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if (index0 < index1) swap(nums, i, index1);//换过去的是1，触发交换 ->好像不是原地排序了，交换之后前面的1跑到后面了
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index0++;
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index1++;
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} else if (nums[i] == 1) {
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swap(nums, i, index1);
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index1++;
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}
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}
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}
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public void swap(int[] nums, int i, int j) {
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int temp = nums[i];
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nums[i] = nums[j];
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nums[j] = temp;
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}
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}
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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(minWindow(s, t));
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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(minWindow1(s, t));
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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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}
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package com.markilue.leecode.hot100.second;
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import java.util.ArrayList;
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import java.util.List;
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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: 2023-04-27 11:55
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*@Description:
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* TODO 力扣78 子集:
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* 给你一个整数数组 nums ，数组中的元素 互不相同 。返回该数组所有可能的子集（幂集）。
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* 解集 不能 包含重复的子集。你可以按 任意顺序 返回解集。
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*@Version: 1.0
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*/
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public class T36_78_Subsets {
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List<Integer> cur = new ArrayList<>();
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List<List<Integer>> result = new ArrayList<>();
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public List<List<Integer>> subsets(int[] nums) {
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backtracking(nums,0);
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return result;
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}
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public void backtracking(int[] nums, int start) {
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result.add(new ArrayList<>(cur));
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for (int i = start; i < nums.length; i++) {
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cur.add(nums[i]);
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backtracking(nums,i+1);
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cur.remove(cur.size()-1);
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}
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}
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}
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