72. Edit Distance

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:

1. Insert a character
2. Delete a character
3. 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')

感觉这题很厉害，DP ， 先领会思想

class Solution {
public:
	int minDistance(string word1, string word2) {
		int len1 = word1.length();
		int len2 = word2.length(); 
		vector<vector<int>> f(len1+1, vector<int>(len2+1, 0));  
		f[0][0] = 0; 
		//  f[i][j]  表示 前 i个数和前j个数
		for (int i = 1; i <= len2; ++i)
			f[0][i] = i;   // 即 word1 为空
		int  temp = 0;
		for (int j = 1; j <= len1; ++j)
			f[j][0] = j;  // 即word2 为空 


		for (int i = 1; i <= len1; ++i)
		{
			for (int j = 1; j <= len2; ++j)
			{
				temp = min(f[i - 1][j] + 1, f[i][j - 1] + 1);
				
				if (word1[i - 1] == word2[j - 1])
				{
					f[i][j] = min(temp, f[i-1][j-1]);
				}
				else f[i][j] = min(temp, f[i - 1][j - 1] + 1);
			}
		}

		return  f[len1][len2];
	}

	int min(int a, int b)
	{
		return  a > b ? b : a;
	}
};