题目

edit-distance

算法

https://segmentfault.com/a/1190000003741294

http://www.cnblogs.com/grandyang/p/4344107.html

$$o(n^2)$$

• dp[i][j]:从word1的前i个字符转换到word2的前j个字符所需步骤

代码

public class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length(), n = word2.length();
        int[][] dp = new int[m + 1][n + 1];
        // 初始化空字符串的情况
        for(int i = 1; i <= m; i++){
            dp[i][0] = i;
        }
        for(int i = 1; i <= n; i++){
            dp[0][i] = i;
        }
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                // 增加操作：str1a变成str2后再加上b，得到str2b
                int insertion = dp[i][j-1] + 1;
                // 删除操作：str1a删除a后，再由str1变为str2b
                int deletion = dp[i-1][j] + 1;
                // 替换操作：先由str1变为str2，然后str1a的a替换为b，得到str2b
                int replace = dp[i-1][j-1] + (word1.charAt(i - 1) == word2.charAt(j - 1) ? 0 : 1);
                // 三者取最小
                dp[i][j] = Math.min(replace, Math.min(insertion, deletion));
            }
        }
        return dp[m][n];
    }
}