LeetCode-72-Edit Distance

算法描述：

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：动态规划题。递推式为：

dp[i][j] = dp[i-1][j-1] if(word1[i-1]==word2[j-1])

dp[i][j] = min(min(dp[i][j-1], dp[i-1][j]), dp[])+1 if(word1[i-1]!=word2[j-1])

注意初始化问题。

    int minDistance(string word1, string word2) {
        vector<vector<int>> dp(word1.size()+1,vector<int>(word2.size()+1,0));
        dp[0][0] = 0;
        for(int i=1; i <= word1.size(); i++) dp[i][0]=i;
        for(int j=1; j <= word2.size(); j++) dp[0][j]=j;
        for(int i=1; i <= word1.size(); i++){
            for(int j=1; j <= word2.size(); j++){
                if(word1[i-1]==word2[j-1]){
                    dp[i][j] = 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;
                }
            }
        }
        return dp[word1.size()][word2.size()];
    }