Leetcode 72. 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

又是字符匹配问题，利用数组dp[i][j]表示word1的前i个字符与word2前j个字符匹配需要的最少操作次数。
状态转移方程：
if word1[i]==word2[j] dp[i][j]=dp[i-1][j-1]
if word1[i]!=word2[j] dp[i][j]=min(dp[i-1][j-1],dp[i-1][j],dp[i][j-1])
代码如下：

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