problem 819. Most Common Word solution: class Solution { public: string mostCommonWord(string paragraph, vector<string>& banned) { unordered_map<string, int> m; string word = ""; ; i<paragraph.size(); ) { string word = "&…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 正则+统计 日期 题目地址:https://leetcode.com/problems/most-common-word/description/ 题目描述 Given a paragraph and a list of banned words, return the most frequent word that is not in the li…

problem 1071. Greatest Common Divisor of Strings solution class Solution { public: string gcdOfStrings(string str1, string str2) { return (str1+str2==str2+str1) ? (str1.substr(, gcd(str1.size(), str2.size()))) : ""; } }; 参考 1. Leetcode_easy_1071…

problem 748. Shortest Completing Word 题意: solution1: class Solution { public: string shortestCompletingWord(string licensePlate, vector<string>& words) { string res = ""; unordered_map<char, int> freq; ; for(char ch : licensePlat…

[SP1812]LCS2 - Longest Common Substring II 题面 洛谷 题解 你首先得会做这题. 然后就其实就很简单了, 你在每一个状态\(i\)打一个标记\(f[i]\)表示状态\(i\)能匹配到最长的子串长度, 显然\(f[i]\)可以上传给\(f[i.fa]\). 然后去每个串和第\(1\)个串\(f\)的最小值的最大值即可. 代码 #include <iostream> #include <cstdio> #include <cstdlib&…

[SP1811]LCS - Longest Common Substring 题面 洛谷 题解 建好后缀自动机后从初始状态沿着现在的边匹配, 如果失配则跳它的后缀链接,因为你跳后缀链接到达的\(Endpos\)集合中的串肯定是当前\(Endpos\)中的后缀,所以这么做是对的. 你感性理解一下,这样显然是最大的是吧... 具体实现看代码: 代码 #include <iostream> #include <cstdio> #include <cstdlib> #inclu…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/ 题目描述 Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in…

Given a paragraph and a list of banned words, return the most frequent word that is not in the list of banned words.  It is guaranteed there is at least one word that isn't banned, and that the answer is unique. Words in the list of banned words are…

problem 720. Longest Word in Dictionary 题意: solution1: BFS; class Solution { public: string longestWord(vector<string>& words) { string res = ""; unordered_set<string> s(words.begin(), words.end()); queue<string> q; for(aut…

Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0. Example…