LeetCode每日一题：edit distance

问题描述

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)
You have the following 3 operations permitted on a word:
a) Insert a character
b) Delete a character
c) Replace a character

问题分析

求编辑距离也是动态规划中比较经典的例子了：
设dp[i][j]为word1的前i个字符和word2的前j个字符的编辑距离
则dp[i][j]=min(dp[i-1][j],dp[i][j-1],dp[i-1][j-1])+1;
三种情况分别对应插入、删除、替换字符，取其中最小的，最后再+1即可

代码实现

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