Edit Distance

最新推荐文章于 2025-01-02 22:18:22 发布

Nereus_Li

最新推荐文章于 2025-01-02 22:18:22 发布

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分类专栏： LeetCode 文章标签： DP string

本文链接：https://blog.youkuaiyun.com/li_chihang/article/details/44803099

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本文介绍了一种计算两个字符串之间编辑距离的算法，并提供了详细的实现步骤。通过插入、删除和替换字符的操作，该算法能够计算出从一个字符串转换为另一个字符串所需的最小操作数。

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Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

#include<iostream>
#include<vector>
#include<string>
using namespace std;

int minDistance(string word1, string word2) {
	int m = word1.size()+1;
	int n = word2.size()+1;

	vector<vector<int>>Distance(m,vector<int>(n,0));

	for (int i = 0;i!=m;++i)
		Distance[i][0] = i;
	for (int j = 0;j!=n;++j)
		Distance[0][j] = j;

	for (int i = 1;i!=m;++i)
	{
		for (int j =1;j!=n;++j)
		{
			if (word1[i-1]==word2[j-1])
				Distance[i][j] = Distance[i-1][j-1];
			else
				Distance[i][j] = Distance[i-1][j-1]+1;
			Distance[i][j] = min(Distance[i][j],min(Distance[i-1][j]+1,Distance[i][j-1]+1));
		}
	}
	return Distance[m-1][n-1];
}