Edit Distance

题目描述：

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

解题思路：

使用动态规划，状态dp[i][j]表示字符串words1的前i个字符构成的子串和字符串words2的前j个字符构成的子串之间的最小编辑距离。
状态转移方程为:
如果words1[i-1]=words2[j-1],那么dp[i][j]=dp[i-1][j-1]
否则dp[i][j]=min(dp[i-1][j-1],dp[i-1][j],dp[i][j-1])+1

AC代码如下：

class Solution {
public:
	int minDistance(string word1, string word2) {
		int n1 = word1.size();
		int n2 = word2.size();
		if (n1 == 0) return n2;
		if (n2 == 0) return n1;
		vector<vector<int>> dp(n1+1, vector<int>(n2+1, 0));
		//initial the first row
		for (int i = 0; i <= n2; ++i) dp[0][i] = i;
		//initial the first col
		for (int i = 0; i <= n1; ++i) dp[i][0] = i;
		for (int i = 1; i <= n1; ++i){
			for (int j = 1; j <= n2; ++j){
				if (word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
				else dp[i][j] = myMin(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1;
			}
		}
		return dp[n1][n2];
	}
private:
	int myMin(int a, int b, int c)
	{
		return a < b ? (a<c?a:c) : (b<c?b:c);
	}
};