72. Edit Distance (String; DP)

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

思路：动态规划。将所要求的min step作为状态，dp[i][j]表示word2的前j各字符通过word1的前i各字符转换最少需要多少步。可以看到有两个以上的string，通常状态要定义为二维数组，表示两个字符串前几个字符之间的关系。

class Solution {
public:
    int minDistance(string word1, string word2) {
        if(word1.length()==0) return word2.length();
        if(word2.length()==0) return word1.length();
        
        vector<vector<int>> dp(word1.length(), vector<int>(word2.length(),0));
        
        if(word1[0]==word2[0]) dp[0][0] = 0;
        else dp[0][0] = 1;
       
        for(int i = 0; i < word1.length(); i++){
            for(int j = 0; j < word2.length(); j++){
                if(i>0 && j>0){
                    if(word1[i]==word2[j]){
                        dp[i][j] =min(min(dp[i-1][j], dp[i][j-1])+1,dp[i-1][j-1]);
                    }
                    else{
                        dp[i][j]=min(min(dp[i-1][j], dp[i][j-1]),dp[i-1][j-1])+1;
                    }
                }
                else if(i>0){ 
                    if(word1[i]==word2[j]){
                        dp[i][j] =i;
                    }
                    else{
                        dp[i][j]=dp[i-1][j]+1;
                    }
                }
                else if(j>0){
                    if(word1[i]==word2[j]){
                        dp[i][j] =j;
                    }
                    else{
                        dp[i][j]=dp[i][j-1]+1;
                    }
                }
            }
        }
        
        return dp[word1.length()-1][word2.length()-1];
    }
};