【leetcode】Edit Distance

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

采用动态规划求解

1. d[0, j] = j;

2. d[i, 0] = i;

3. d[i, j] = d[i-1, j - 1] if A[i] == B[j]

4. d[i, j] = min(d[i-1, j - 1], d[i, j - 1], d[i-1, j]) + 1  if A[i] != B[j]

 

class Solution {
public:
    int minDistance(string word1, string word2) {
       
        int n1=word1.length();
        int n2=word2.length();
       
        if(n1==0) return n2;
        if(n2==0) return n1;
 
        //采用二维数组效率更高 
        //vector<vector<int> > dp(n1+1,vector<int>(n2+1));
       
        int **dp=new int*[n1+1];
        for(int i=0;i<n1+1;i++)
        {
            dp[i]=new int[n2+1];
        }
       
       
        dp[0][0]=0;
        for(int i=1;i<=n1;i++)
        {
            dp[i][0]=i;
        }
        for(int j=1;j<=n2;j++)
        {
            dp[0][j]=j;
        }
       
        for(int i=1;i<=n1;i++)
        {
            for(int j=1;j<=n2;j++)
            {
               
                if(word1[i-1]==word2[j-1])
                {
                    dp[i][j]=dp[i-1][j-1];
                }
                else
                {
                    dp[i][j]=min(dp[i-1][j],min(dp[i][j-1],dp[i-1][j-1]))+1;
                }
            }
        }
       
        int result=dp[n1][n2];
        for(int i=0;i<n1+1;i++)
        {
            delete[] dp[i];
        }
       
        return result;
    }
};