72. Edit Distance

Posted 2020-11-10 妖域大都督

问题描述：

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

1. Insert a character
2. Delete a character
3. Replace a character

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace ‘h‘ with ‘r‘)
rorse -> rose (remove ‘r‘)
rose -> ros (remove ‘e‘)

Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove ‘t‘)
inention -> enention (replace ‘i‘ with ‘e‘)
enention -> exention (replace ‘n‘ with ‘x‘)
exention -> exection (replace ‘n‘ with ‘c‘)
exection -> execution (insert ‘u‘)

解题思路：

这道题可以用动态规划来做：

dp[i][j]表示的是word1的前i个字符变成word2的前j个字符的最少操作

当word1为空的时候，变成word2所需要的操作为word2.size()所以i = 0时dp[i][j] = j

同理： j = 0时 dp[i][j] = i

状态转移方程：

word1[i] = word2[j] : dp[i][j] = dp[i-1][j-1]

else: dp[i][j] = min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1])) + 1 

 

代码：

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