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Leetcode 583. 兩個字符串的刪除操作

題目：583. 兩個字符串的刪除操作
解析：代碼隨想錄解析

解題思路

dp數組的含義是，從word1從0到i-1，word2從0到j-1匹配上最少需要刪除多少個字符

代碼

class Solution {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 = 0; i <= m; i++)dp[i][0] = i;for (int j = 0; 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];elsedp[i][j] = Math.min(dp[i][j-1] + 1 , Math.min(dp[i-1][j] + 1, dp[i-1][j-1] + 2));}}return dp[m][n];}
}

總結

暫無

Leetcode 72. 編輯距離

題目：72. 編輯距離
解析：代碼隨想錄解析

解題思路

和上面那題的區別是，改了一個dp[i-1][i-1] + 1，因為可以修改，而上一題只能刪除兩個

代碼

class Solution {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 = 0; i <= m; i++)dp[i][0] = i;for (int j = 0; 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];elsedp[i][j] = Math.min(dp[i-1][j], Math.min(dp[i][j-1], dp[i-1][j-1])) + 1;}}return dp[m][n];}
}

總結

暫無