leetcode 72 Edit Distance c++_edit distance dynamic programmingc++-优快云博客

博客围绕编辑距离问题展开，给定两个单词，求将一个单词转换为另一个所需的最少操作数，允许插入、删除、替换操作。采用动态规划求解，定义dp[i][j]表示特定转换的最小步骤数，通过选三种操作中的最小步骤解决问题，最后还提及空间优化。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

72. Edit Distance

Hard

197930FavoriteShare

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:

Insert a character
Delete a character
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')

这道题很明显的动态规划，不过重点是要想清楚什么时候用replace，insert和delete；

dp[i][j]表示将word1前i个字符转化成word2前j个字符所需要的最小步骤数；

一开始我一直在想到底什么时候用哪种，后来发现只要选三种中的最小哪种就解决了；

dp[i][j] = min(dp[i-1][j-1]（替换）,min(dp[i-1][j](word1前i-1个字符就可以转化成word2前j个字符，删除),dp[i][j-1](word1前i个字符只能转换成word2前j-1个字符，插入word[j-1]）);

最后做一下空间优化。

class Solution {
public:
    int minDistance(string word1, string word2) 
    {
        int len1 = word1.size();
        int len2 = word2.size();
        vector<int> dp(len2+1,0);
        if(len1 == 0) return len2;
        if(len2 == 0) return len1;
        for(int j = 0;j <= len2; ++j) dp[j] = j;
        for(int i = 1; i <= len1; ++i)
        {
            int pre = i-1;
            dp[0] = i;
            for(int j=1; j<= len2 ;++j)
            {
                int cur = dp[j];
                if(word1[i-1] == word2[j-1]) dp[j] = pre;
                else dp[j] = min(pre,min(dp[j-1],dp[j])) + 1;
                pre = cur;
            }
        }
        return dp[len2];
    }
};