72. Edit Distance

标签：distance   min   思路   rac   mit   经典   ted   字符   ...   

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：经典dp题。用dp[i][j]表示word1[0...i]与word2[0...j]的最小编辑距离。

把word1和word2相关字符对齐后，只要看最后一位是否一样即可

1.word1为空，word2为空，此时无意义

2.word1为空，word2不为空，word1需要插入操作  //dp[i][j]=dp[i][j-1]+1

3.word1不为空，word2为空，word1需要删除操作  //dp[i][j]=dp[i-1][j]+1

4.word1不为空，word2不为空，如果word1==word2，此时不操作，否则把word1修改为word2 //dp[i][j]=word1==word2?dp[i-1][j-1]:dp[i-1][j-1]+1

class Solution {
public:
    int minDistance(string word1, string word2) {
        //dp[i][j] means s[0...i] to t[0...j]
        int n=word1.length(),m=word2.length();
        int dp[n+1][m+1];
        for(int i=0;i<=n;i++)
            dp[i][0]=i;
        for(int i=0;i<=m;i++)
            dp[0][i]=i;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++){
                if(word1[i-1]==word2[j-1]){
                    dp[i][j]=dp[i-1][j-1];
                }
                else {
                    dp[i][j]=1+min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1]));
                }
            }
        return dp[n][m];
    }
};

72. Edit Distance

标签：distance   min   思路   rac   mit   经典   ted   字符   ...   

原文地址：http://www.cnblogs.com/tsunami-lj/p/6534382.html