给你两个单词 word1 和 word2，请你计算出将 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：

1、插入一个字符
2、删除一个字符
3、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’)

#include <iostream>
#include <algorithm>
#include <windows.h>
using namespace std;

int minDistance(string word1, string word2) {
	int** dp = new int*[word1.length() + 1];
	for (int i = 0; i <= word1.length(); i++)
		dp[i] = new int[word2.length() + 1];
	for (int i = 0; i <= word1.length(); i++)
		dp[i][0] = i;
	for (int j = 0; j <= word2.length(); j++)
		dp[0][j] = j;
	for (int i = 1; i <= word1.length(); i++)
		for (int j = 1; j <= word2.length(); j++) {
			if (word1[i - 1] == word2[j - 1])
				dp[i][j] = dp[i - 1][j - 1];
			else {
				int temp = min(dp[i - 1][j], dp[i][j - 1]);
				dp[i][j] = min(temp, dp[i - 1][j - 1]) + 1;
			}
		}
	return dp[word1.length()][word2.length()];
}

int main() {
	string word1 = "";
	string word2 = "a";
	cout << minDistance(word1, word2) << endl;
	system("pause");
	return 0;
}