代码随想录算法训练营第45天

115.不同的子序列

但相对于刚讲过 392.判断子序列，本题 就有难度了 ，感受一下本题和 392.判断子序列 的区别。

代码随想录

class Solution {public int numDistinct(String s, String t) {int lenS = s.length();int lenT = t.length();int[][] dp = new int[lenS + 1][lenT + 1];for (int i = 0; i <= lenS; i++) {dp[i][0] = 1; // An empty string t has one way to be a subsequence of any string s}for (int i = 1; i <= lenS; i++) {for (int j = 1; j <= lenT; j++) {if (s.charAt(i - 1) == t.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1];} else {dp[i][j] = dp[i][j - 1];}}}return dp[lenS][lenT];}
}

583. 两个字符串的删除操作

本题和动态规划：115.不同的子序列 相比，其实就是两个字符串都可以删除了，情况虽说复杂一些，但整体思路是不变的。

代码随想录

class Solution {public int minDistance(String word1, String word2) {int len1 = word1.length();int len2 = word2.length();int[][] dp = new int[len1 + 1][len2 + 1];for (int i = 0; i <= len1; i++) {dp[i][0] = i;}for (int j = 0; j <= len2; j++) {dp[0][j] = j;}for (int i = 1; i <= len1; i++) {for (int j = 1; j <= len2; j++) {if (word1.charAt(i - 1) == word2.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1];} else {dp[i][j] = Math.min(dp[i - 1][j - 1] + 2,Math.min(dp[i - 1][j] + 1, dp[i][j - 1] + 1));}}}return dp[len1][len2];}
}

72. 编辑距离

最终我们迎来了编辑距离这道题目，之前安排题目都是为了 编辑距离做铺垫。

代码随想录

public int minDistance(String word1, String word2) {int m = word1.length();int n = word2.length();int[][] dp = new int[m + 1][n + 1];for (int i = 1; i <= m; i++) {dp[i][0] =  i;}for (int j = 1; j <= n; j++) {dp[0][j] = j;}for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {if (word1.charAt(i - 1) == word2.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1];} else {dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1], dp[i][j - 1]), dp[i - 1][j]) + 1;}}}return dp[m][n];
}

编辑距离总结篇

做一个总结吧

代码随想录