代码随想录算法训练营第五十六天| 583. 两个字符串的删除操作 72. 编辑距离

代码随想录算法训练营第五十六天| 583. 两个字符串的删除操作 72. 编辑距离

一、力扣583. 两个字符串的删除操作

题目链接
思路：相等时不删除，不相等时，两个字符串各删除一个，比大小，删除用步骤少的。

class Solution {public int minDistance(String word1, String word2) {int[][] dp = new int[word1.length()+1][word2.length()+1];for (int i = 0; i <= word1.length(); i++) {dp[i][0] = i;}for (int i = 0; i <= word2.length(); i++) {dp[0][i] = i;}for (int i = 1; i <= word1.length(); i++) {for (int j = 1; j <= word2.length(); 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, dp[i][j-1] + 1);}}}return dp[word1.length()][word2.length()];}
}

二、力扣72. 编辑距离

题目链接
思路：区间结尾不相等，有增删换，dp[i][j] = min({dp[i - 1][j - 1], dp[me - 1][j], dp[me][j - 1]}) + 1

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