算法训练营第五十六天|583. 两个字符串的删除操作 72. 编辑距离

本文主要是介绍算法训练营第五十六天|583. 两个字符串的删除操作 72. 编辑距离，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Leetcode583. 两个字符串的删除操作

文章链接：代码随想录
题目链接：583. 两个字符串的删除操作

思路：直接记录需要改（增或删）几个，也就是求不公共的子序列

class Solution {
public:int minDistance(string word1, string word2) {vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;for (int i = 1; i <= word1.size(); i++){for (int j = 1; j <= word2.size(); j++){if (word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1];else dp[i][j] = min(dp[i - 1][j - 1] + 2, min(dp[i][j - 1] + 1, dp[i - 1][j] + 1));}}return dp[word1.size()][word2.size()];}
};

也可以记录最长公共子序列，再减

class Solution {
public:int minDistance(string word1, string word2) {vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));for (int i = 1; i <= word1.size(); i++){for (int j = 1; j <= word2.size(); j++){if (word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;else dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);}}return word1.size() + word2.size() - dp[word1.size()][word2.size()] * 2;}
};

Leetcode72. 编辑距离

文章链接：代码随想录
题目链接：72. 编辑距离

思路：和上一题相比，差别在于多了替换，因此dp[i - 1][j - 1] 只需要多加一步即可变为dp[i][j]。

class Solution {
public:int minDistance(string word1, string word2) {vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;for (int j = 1; j <= word2.size(); j++) dp[0][j] = j;for (int i = 1; i <= word1.size(); i++){for (int j = 1; j <= word2.size(); j++){if (word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1];else dp[i][j] = min(dp[i - 1][j - 1] + 1, min(dp[i - 1][j] + 1, dp[i][j - 1] + 1));}}return dp[word1.size()][word2.size()];}
};