leecode更新
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markilue
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package com.markilue.leecode.dynamic;
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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.dynamic
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*@Author: dingjiawen
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*@CreateTime: 2022-12-20 10:17
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*@Description:
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* TODO 力扣583题 两个字符串的删除操作:
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* 给定两个单词 word1 和 word2 ，返回使得 word1 和 word2 相同所需的最小步数。
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* 每步 可以删除任意一个字符串中的一个字符。
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*@Version: 1.0
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*/
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public class T35_MinDistance {
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@Test
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public void test(){
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String word1 = "sea";
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String word2 = "eat";
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System.out.println(minDistance1(word1,word2));
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}
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@Test
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public void test1(){
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String word1 = "leetcode";
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String word2 = "etco";
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System.out.println(minDistance1(word1,word2));
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}
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/**
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* 思路:本质上还是求最少删除多少个字符会相等
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* TODO 动态规划法:
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* (1)dp定义:dp[i][j]表示使用word1[0-i]和word2[0-j]需要删除多少个变成一样
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* (2)dp状态转移方程:
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* 1.如果两个数相等：和以前没加这两个数时一样
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* if word1[i]==word2[j] dp[i][j]=dp[i-1][j-1]
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* 2.如果两个数不相等：可以去掉当前这个word1[i]或者去掉word1[j]看谁减去的字符多
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* else dp[i][j]=min(dp[i-1][j]+1,dp[i][j-1]+1)
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* (3)dp初始化:dp[i][0]=i;dp[0][i]=i
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* (4)dp遍历顺序:
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* (5)dp举例推导：以word1 = "sea", word2 = "eat"为例
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* [0 e a t]
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* i=0: 0 1 2 3
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* i=s: 1 2 3 4
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* i=e: 2 1 2 3
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* i=a: 3 2 1 2
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* 速度击败62.85%，内存击败66.73%
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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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//初始化
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for (int i = 0; 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[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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for (int j = 1; j < dp[0].length; j++) {
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if (word1.charAt(i-1)==word2.charAt(j-1)) dp[i][j]=dp[i-1][j-1];
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else dp[i][j]=Math.min(dp[i-1][j]+1,dp[i][j-1]+1);
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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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/**
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* 代码随想录另一种思路：这题和求最长公共子序列类似：
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* 只要求出两个字符串的最长公共子序列长度即可，那么除了最长公共子序列之外的字符都是必须删除的，
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* 最后用两个字符串的总长度减去两个最长公共子序列的长度就是删除的最少步数。
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* 速度击败62.85%，内存击败14.5%
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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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int[][] dp = new int[word1.length() + 1][word2.length() + 1];
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//初始化
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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)) dp[i][j]=dp[i-1][j-1]+1;
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else dp[i][j]=Math.max(dp[i-1][j],dp[i][j-1]);
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}
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}
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return word1.length()+word2.length()-dp[word1.length()][word2.length()]*2;
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}
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/**
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* 官方题解中最快：本质上是最长子序列的解法
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* 速度击败100%，内存击败91.88% 3ms
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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 minDistance2(String word1, String word2) {
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int m = word1.length();
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int n = word2.length();
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if (m < n) {
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String s = word1;
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word1 = word2;
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word2 = s;
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int t = m;
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m = n;
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n = t;
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}
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char[] w1 = word1.toCharArray();
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char[] w2 = word2.toCharArray();
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int[] prev = new int[n + 1];
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int[] curr = new int[n + 1];
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for (char c : w1) {
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for (int j = 0; j < n; j++) {
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if (c == w2[j]) {
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curr[j + 1] = prev[j] + 1;
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} else {
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curr[j + 1] = Math.max(prev[j + 1], curr[j]);
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}
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}
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int[] t = prev;
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prev = curr;
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curr = t;
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}
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return m + n - 2 * prev[n];
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}
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}
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@ -0,0 +1,132 @@
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package com.markilue.leecode.dynamic;
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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.dynamic
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*@Author: dingjiawen
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*@CreateTime: 2022-12-20 11:36
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*@Description:
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* TODO 力扣72题 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 T36_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(minDistance(word1,word2));
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}
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@Test
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public void test1(){
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String word1 = "intention";
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String word2 = "execution";
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System.out.println(minDistance(word1,word2));
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}
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/**
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* 由于可以插入，可以替换，可以删除，本人没有明确的想法，这里先用距离推导状态转移方程
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* TODO 动态规划法:
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* 1.dp定义：dp[i][j]表示word1[0-i]变成word2[0-j]所需最少步骤
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* 2.dp状态转移方程:
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* 1.当word1[i]=word2[j]时,那么仅需要处理i,j之前的数
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* dp[i][j]=dp[i-1][j-1]
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* 2.当word1[i]!=word2[j]时
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* 1)如果word1长度不够，他可能需要从word[j]添加一个数 dp[i][j-1]+1
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* 2)如果word1长度刚好,他可能需要在word[i-1][j-1]的基础上替换当前这个数 dp[i-1][j-1]+1
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* 3)如果word1长度超过，他直接把他删了,在dp[i-1][j]的基础上删除 dp[i-1][j]+1
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* dp[i][j]=min(dp[i][j-1]+1,dp[i-1][j-1]+1,dp[i-1][j]+1)
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* 3.dp初始化:dp[0][i]=i;dp[i][0]=i
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* 4.dp遍历顺序:
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* 5.dp举例推导:以word1 = "horse", word2 = "ros"为例
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* [0 r o s]
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* i=0: 0 1 2 3
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* i=h: 1 1 2 3
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* i=o: 2 2 1 2
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* i=r: 3 2 2 2
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* i=s: 4 3 3 2
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* i=e: 5 4 4 3
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* 速度击败95.89%，内存击败5.17%
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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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char[] char1 = word1.toCharArray();
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char[] char2 = word2.toCharArray();
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int length1 = char1.length;
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int length2 = char2.length;
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int[][] dp = new int[length1 + 1][length2 + 1];
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//初始化
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for (int i = 0; i < dp.length; i++) {
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dp[i][0]=i;
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}
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for (int i = 0; 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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for (int j = 1; j < dp[0].length; j++) {
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if(char1[i-1]==char2[j-1]) dp[i][j]=dp[i-1][j-1];
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else dp[i][j]=Math.min(Math.min(dp[i][j-1]+1,dp[i-1][j-1]+1),dp[i-1][j]+1);
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}
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}
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return dp[length1][length2];
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}
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/**
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* 官方题解中最快的方法，本质上还是dp,使用递归实现
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* 速度击败100%，内存击败5.17% 2ms
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*/
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int[][] meno ;
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public int minDistance1(String word1, String word2){
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meno = new int[word1.length()][word2.length()];
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for (int[] ints : meno) {
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Arrays.fill(ints,-1);
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}
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return dp(word1,word1.length()-1,word2,word2.length()-1);
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}
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private int dp(String word1, int i, String word2, int j) {
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if(i==-1){
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//i不够了,全添加到j
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return j+1;
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}
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if(j==-1){
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//j不够了,把i全删了
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return i+1;
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}
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if(meno[i][j]!=-1){
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return meno[i][j];//以前算过了，直接返回；有点像记忆化搜索
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}
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//状态转移方程
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if(word1.charAt(i)==word2.charAt(j)){
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meno[i][j] = dp(word1,i-1,word2,j-1);
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}else{
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meno[i][j] = Math.min(Math.min(dp(word1,i,word2,j-1),dp(word1,i-1,word2,j)),dp(word1,i-1,word2,j-1))+1;
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}
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return meno[i][j];
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}
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}
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