LeetCode75——编辑距离

题目：

给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。

你可以对一个单词进行如下三种操作：

插入一个字符
删除一个字符
替换一个字符
示例 1:

输入: word1 = "horse", word2 = "ros"
输出: 3
解释: 
horse -> rorse (将 'h' 替换为 'r')
rorse -> rose (删除 'r')
rose -> ros (删除 'e')
示例 2:

输入: word1 = "intention", word2 = "execution"
输出: 5
解释: 
intention -> inention (删除 't')
inention -> enention (将 'i' 替换为 'e')
enention -> exention (将 'n' 替换为 'x')
exention -> exection (将 'n' 替换为 'c')
exection -> execution (插入 'u')

思路：

见下面的文章，讲解比较清楚

https://leetcode-cn.com/problems/edit-distance/solution/bian-ji-ju-chi-mian-shi-ti-xiang-jie-by-labuladong/

代码：

class Solution {
public:
	int minDistance(string s1, string s2)
	{
		int m = s1.length(), n = s2.length();
		vector<vector<int>>dp(m + 1, vector<int>(n + 1));
		for (int i = 1; i <= m; i++)
			dp[i][0] = i;
		for (int j = 1; j <= n; j++)
			dp[0][j] = j;

		for (int i = 1; i <= m; i++) {
			for (int j = 1; j <= n; j++)
			{
				if (s1[i - 1] == s2[j - 1])
					dp[i][j] = dp[i - 1][j - 1];
				else
					dp[i][j] = min_three(dp[i - 1][j] + 1, dp[i][j - 1] + 1, dp[i - 1][j - 1] + 1);
			}

		}

		return dp[m][n];

	}

	int min_three(int a, int b, int c)
	{
		return min(a, min(b, c));
	}
};