72. Edit Distance

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分类专栏： LeetCode

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本文深入解析编辑距离算法，一种衡量两个字符串相似度的方法。通过插入、删除或替换字符操作，计算转换一个单词到另一个单词所需的最少步骤。文章提供了一个C++实现的例子，详细解释了动态规划方法的应用。

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72. Edit Distance

Hard

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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')

Accepted

207,049

Submissions

515,738

参考https://www.cnblogs.com/zle1992/p/8872954.html

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