Leetcode dp Edit Distance

最新推荐文章于 2022-02-06 11:01:06 发布

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最新推荐文章于 2022-02-06 11:01:06 发布

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本文介绍了一个经典的字符串处理问题——编辑距离问题，通过动态规划的方法来解决两个字符串之间的转换步骤数。详细阐述了如何通过增、删、替换单个字符的操作，计算从一个字符串转换到另一个字符串所需的最少操作次数。

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Edit Distance

Total Accepted: 14997 Total Submissions: 59129 My Submissions

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

题意：有两个单词，可以选择以下三种操作：增、删、替换一个字符，问需要经过多少次操作才能把第一个单词变为第二个
思路：dp
f[i][j] 表示word1[1..j]到word2[1..i]的最小编辑距离
if word2[i] == word1[j], f[i][j] = f[i - 1][j - 1]
else f[i][j] = min{f[i - 1][j - 1], f[i][j - 1], f[i - 1][j]} + 1
复杂度：时间O(n^2)，空间O(n^2)

int minDistance(string word1, string word2) {
	if(word1.empty()) return word2.size();
	if(word2.empty()) return word1.size();


	vector<vector<int> > dp(word1.size() + 1, vector<int>(word2.size() + 1, 0)); //
	//初始化
	for(int i = 0; i <= word1.size(); ++i) dp[i][0] = i;
	for(int j = 0; j <= word2.size(); ++j) dp[0][j] = j;


	//迭代
	for(int i = 0; i < word1.size(); ++i){
		for(int j = 0; j < word2.size(); ++j){
			int ii  = i + 1, jj = j + 1;
			if(word1[i] == word2[j]) dp[ii][jj] = dp[ii - 1][jj - 1];
			else dp[ii][jj] = min(min(dp[ii][jj - 1], dp[ii - 1][jj]), dp[ii - 1][jj - 1]) + 1;
			
		}
	}
	return dp[word1.size()][word2.size()];
}