Leetcode:72. Edit Distance

Description

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

思路

  • 动态规划,dp[i][j]表示words1[0..i]和words2[0..j]最小的编辑距离是多少
  • dp[i][j] =min( min(dp[i - 1][j - 1] + count, dp[i - 1][j] + 1), dp[i][j - 1] + 1)
  • 其中当words1[i]==words2[j]时,count = 1; 否则,count = 0;

代码

class Solution {
public:
  int minDistance(string word1, string word2) {
		int len1 = word1.size(), len2 = word2.size();
		if (len1 == 0) return len2;
		if (len2 == 0) return len1;

		vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1, 0));

		dp[0][0] = 0;
		for (int i = 1; i <= len1; ++i)
			dp[i][0] = i;
			
		for (int j = 1; j <= len2; ++j)
			dp[0][j] = j;
			
		for (int i = 1; i <= len1; ++i){
			for (int j = 1; j <= len2; ++j){
				int count = (word1[i - 1] == word2[j - 1] ? 0 : 1);
				dp[i][j] = min(min(dp[i - 1][j - 1] + count, dp[i - 1][j] + 1), dp[i][j - 1] + 1) ;
			}
		}

		return dp[len1][len2];

	}
};
原文地址:https://www.cnblogs.com/lengender-12/p/6921133.html