LeetCode 914卡盘分组】的更多相关文章

In a deck of cards, each card has an integer written on it. Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where: Each group has exactly X cards. All the cards in eac…
原题 题目原意可转换为 两组有大于等于2的公因数 /** * @param {number[]} deck * @return {boolean} */ var hasGroupsSizeX = function(deck) { var map = {}; for (let i = 0; i < deck.length; i++) { if (map[deck[i]]) map[deck[i]] += 1; else map[deck[i]] = 1; } var min = map[deck[…
题目 给定一副牌,每张牌上都写着一个整数. 此时,你需要选定一个数字 X,使我们可以将整副牌按下述规则分成 1 组或更多组: 每组都有 X 张牌. 组内所有的牌上都写着相同的整数. 仅当你可选的 X >= 2 时返回 true. 示例 1: 输入:[1,2,3,4,4,3,2,1] 输出:true 解释:可行的分组是 [1,1],[2,2],[3,3],[4,4] 示例 2: 输入:[1,1,1,2,2,2,3,3] 输出:false 解释:没有满足要求的分组. 示例 3: 输入:[1] 输出:…
给定一副牌,每张牌上都写着一个整数. 此时,你需要选定一个数字 X,使我们可以将整副牌按下述规则分成 1 组或更多组: 每组都有 X 张牌. 组内所有的牌上都写着相同的整数. 仅当你可选的 X >= 2 时返回 true. 示例 1: 输入:[1,2,3,4,4,3,2,1] 输出:true 解释:可行的分组是 [1,1],[2,2],[3,3],[4,4] 示例 2: 输入:[1,1,1,2,2,2,3,3] 输出:false 解释:没有满足要求的分组. 示例 3: 输入:[1] 输出:fal…
题目链接: https://leetcode.com/problems/anagrams/?tab=Description   Problem:给一个字符串数组,将其中的每个字符串进行分组,要求每个分组中的各个字符串所包含的字母都相同对字母的前后顺序没有要求!       使用 Map<String, List<String>> map 进行求解   给定的字符串数组为 String[] strs 1.当strs为空时返回new ArrayList<List<Strin…
LeetCode:字母异位词分组[16] 题目描述 给定一个字符串数组,将字母异位词组合在一起.字母异位词指字母相同,但排列不同的字符串. 示例: 输入: ["eat", "tea", "tan", "ate", "nat", "bat"], 输出: [ ["ate","eat","tea"], ["nat"…
LeetCode 49: 字母异位词分组 Group Anagrams 题目: 给定一个字符串数组,将字母异位词组合在一起.字母异位词指字母相同,但排列不同的字符串. Given an array of strings, group anagrams together. 示例: 输入: ["eat", "tea", "tan", "ate", "nat", "bat"], 输出: [ […
Given an array of strings, group anagrams together. For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"], Return: [ ["ate", "eat","tea"], ["nat",…
Given a string, we can "shift" each of its letter to its successive letter, for example: "abc" -> "bcd". We can keep "shifting" which forms the sequence: "abc" -> "bcd" -> ... -> &quo…
914. Flip Game https://www.cnblogs.com/grandyang/p/5224896.html 从前到后遍历,遇到连续两个'+',就将两个加号变成'-'组成新的字符串加入到结果中. class Solution { public: vector<string> generatePossibleNextMoves(string &s) { // write your code here vector<string> result; ;i <…