代码随想录算法训练营第五十五天|583. 两个字符串的删除操作、72. 编辑距离

583. 两个字符串的删除操作

• 刷题https://leetcode.cn/problems/delete-operation-for-two-strings/description/
• 文章讲解https://programmercarl.com/0583.%E4%B8%A4%E4%B8%AA%E5%AD%97%E7%AC%A6%E4%B8%B2%E7%9A%84%E5%88%A0%E9%99%A4%E6%93%8D%E4%BD%9C.html
• 视频讲解https://www.bilibili.com/video/BV1we4y157wB/?vd_source=af4853e80f89e28094a5fe1e220d9062

• 题解：

class Solution {public int minDistance(String word1, String word2) {char[] char1 = word1.toCharArray();char[] char2 = word2.toCharArray();int len1 = char1.length;int len2 = char2.length;int dp[][] = new int [len1 + 1][len2 + 1];for(int i = 1; i <= len1; i++){for(int j = 1; j <= len2; j++){if(char1[i - 1] == char2[j - 1])dp[i][j] = dp[i - 1][j - 1] + 1;elsedp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);}}return len1 + len2 - (2 * dp[len1][len2]);//和leetcode 1143只差在这一行。}
}

72. 编辑距离

• 刷题https://leetcode.cn/problems/edit-distance/description/
• 文章讲解https://programmercarl.com/0072.%E7%BC%96%E8%BE%91%E8%B7%9D%E7%A6%BB.html
• 视频讲解https://www.bilibili.com/video/BV1qv4y1q78f/?vd_source=af4853e80f89e28094a5fe1e220d9062

• 题解：

public int minDistance(String word1, String word2) {int m = word1.length();int n = word2.length();int[][] dp = new int[m + 1][n + 1];// 初始化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++) {// 因为dp数组有效位从1开始// 所以当前遍历到的字符串的位置为i-1 | j-1if (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];
}