Leetcode之Edit Distance

题目：

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')

代码：

-方法1（vector）

class Solution {
public:
    int minDistance(string word1, string word2) {
       int len1 = word1.length();
	int len2 = word2.length();
	vector <vector<int>> res(len1 + 1, vector<int>(len2 + 1, 0));
	
	for (int i = 1; i <= len1; i++) { res[i][0] = i; }
	for (int j = 1; j <= len2; j++) { res[0][j] = j; }

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

			if (min > res[i - 1][j] + 1) min = res[i - 1][j] + 1;
			if (min > res[i][j - 1] + 1) min = res[i][j - 1] + 1;

			res[i][j] = min;
		}
	}

	return res[len1][len2]; 
    }
};

方法二（动态数组）：

class Solution {
public:
    int minDistance(string word1, string word2) {
       	int len1 = word1.length();
	int len2 = word2.length();
	
	int** res = new int*[len1 + 1];
	for (int i = 0; i <= len1; i++) {
		res[i] = new int[len2 + 1];
	}
	res[0][0] = 0;
	for (int i = 1; i <= len1; i++) { res[i][0] = i; }
	for (int j = 1; j <= len2; j++) { res[0][j] = j; }

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

			if (min > res[i - 1][j] + 1) min = res[i - 1][j] + 1;
			if (min > res[i][j - 1] + 1) min = res[i][j - 1] + 1;

			res[i][j] = min;
		}
	}

	int result = res[len1][len2];

	delete []res;
	res = NULL;
	return result;
    }
};