72. Edit Distance 编辑距离

题目：

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

 /*
    解题思路：采用动态规划，Word1和Word2从空串开始，所以dp[i][j]的长度为word1.length()+1、word2.length()+2
             首先初始化第一行和第一列，由于插入。修改、删除的代价都是1，所以第一行第一列dp[0][i]=i,dp[i][0]=i
             递推过程，如果s1[i]==s2[j],则不用修改，其dp[i][j]的值为的dp[i-1][j-1];
             如果s1[i]!=s2[j],则判断dp[i-1][j]+1,表示插入字符
                                   dp[i][j-1]+1，表示删除字符
                                   dp[i-1][j-1]+1表示修改字符
            谁比较小，选取较小的值作为dp[i][j]的值
    */
    public int minDistance(String word1, String word2) {
        if(word1.length()==0){
            return word2.length();
        }
        if(word2.length()==0){
            return word1.length();
        }
        char[] s1=word1.toCharArray();
        char[] s2=word2.toCharArray();
        int[][] dp=new int[s1.length+1][s2.length+1];
        //初始化第一行
        for(int i=0;i<=s2.length;i++){
            dp[0][i]=i;
        }
        //初始化第一列
        for(int i=0;i<=s1.length;i++){
            dp[i][0]=i;
        }
        for(int i=1;i<=s1.length;i++){
            for(int j=1;j<=s2.length;j++){
                if(s1[i-1]==s2[j-1]){
                    dp[i][j]=dp[i-1][j-1];
                }else{
                   int temp=Math.min(dp[i-1][j]+1,dp[i][j-1]+1);
                    dp[i][j]=Math.min(temp,dp[i-1][j-1]+1);
                }
            }//for
        }//for
        return dp[s1.length][s2.length];
    }