LeetCode-72-Edit Distance

标签：leetcode   exp   win   output   初始化   code   问题   exe   exec   

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()];
    }

LeetCode-72-Edit Distance

标签：leetcode   exp   win   output   初始化   code   问题   exe   exec   

原文地址：https://www.cnblogs.com/nobodywang/p/10371614.html