72. Edit Distance

标签：let   with   distance   exe   问题   exp   Plan   rac   code   

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];
    }
};

72. Edit Distance

标签：let   with   distance   exe   问题   exp   Plan   rac   code   

原文地址：https://www.cnblogs.com/yaoyudadudu/p/9128182.html