Edit Distance


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


class Solution {
public:

/*
超时
    int minDistance(string word1, string word2) {
        
        int len1,len2;
        len1=word1.size();
        len2=word2.size();
        return solve(word1,0,len1,word2,0,len2);
    }
    
    int solve(string word1,int index1,int len1,string word2,int index2,int len2)
    {
        if(index1>=len1)
        {
            if(index2>=len2)
                return 0;
            else
                return len2-index2;
        }
        
        if(index2>=len2)
        {
            if(index1>=len1)
                return 0;
            else
                return len1-index1;
        }
        
        if(word1[index1]==word2[index2])
            return solve(word1,index1+1,len1,word2,index2+1,len2);
        else
        {
            int t1=solve(word1,index1+1,len1,word2,index2,len2);
            int t2=solve(word1,index1,len1,word2,index2+1,len2);
            int t3=solve(word1,index1+1,len1,word2,index2+1,len2);
            return min(min(t1,t2),t3)+1;
        }
    }*/
    
    int minDistance(string word1, string word2) {
    
        int len1=word1.size()+1,len2=word2.size()+1;
        vector< vector<int>> dp(len1,vector<int>(len2));
        
        for(int i=0;i<len1;i++)//与空串的距离
            dp[i][0]=i;
        for(int i=0;i<len2;i++)
            dp[0][i]=i;
            
        for(int i=1;i<len1;i++)
            for(int j=1;j<len2;j++)
            {
                if(word1[i-1]==word2[j-1])
                    dp[i][j]=dp[i-1][j-1];
                else
                    dp[i][j]=min(dp[i][j-1],min(dp[i-1][j],dp[i-1][j-1]))+1;
            }
        return dp[len1-1][len2-1];
    }
};