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[…

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…

problem 914. X of a Kind in a Deck of Cards 题意:每个数字对应的数目可以均分为多组含有K个相同数目该数字的数组. 思路:使用 map 结构记录数组中每个元素出现的次数,该题转化为求次数的最大公约数,若所有次数的最大公约数大于或者等于 2,返回 true,否则返回 false. solution: class Solution { public: bool hasGroupsSizeX(vector<int>& deck) { //Greate…

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 遍历 最大公约数 日期 题目地址: https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/description/ 题目描述 In a deck of cards, each card has an integer written on it. Return true if and…

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…

这是悦乐书的第352次更新,第377篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第214题(顺位题号是914).在一副牌中,每张牌上都写有一个整数. 当且仅当您可以选择X >= 2时才返回true,以便可以将整个牌组分成一组或多组牌,其中: 每组都有X张牌. 每组中的所有牌都具有相同的整数. 例如: 输入:[1,2,3,4,4,3,2,1] 输出:true 说明:可能的分区[1,1],[2,2],[3,3],[4,4] 输入:[1,1,1,2,2,2,3,3] 输…

66. Plus One Given a non-negative integer represented as a non-empty array of digits, plus one to the integer. You may assume the integer do not contain any leading zero, except the number 0 itself. The digits are stored such that the most significan…

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…

要求:此题正好和Maximum Depth of Binary Tree一题是相反的,即寻找二叉树的最小的深度值:从根节点到最近的叶子节点的距离. 结题思路:和找最大距离不同之处在于:找最小距离要注意(l<r)? l+1:r+1的区别应用,因为可能存在左右子树为空的情况,此时值就为0,但显然值是不为0的(只有当二叉树为空才为0),所以,在这里注意一下即可! 代码如下: struct TreeNode { int val; TreeNode *left; TreeNode *right; Tree…

题目:Binay Tree Level Order Traversal Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree {,,,#,#,,}, 如下一棵树 转换之后需要输出这样的形式: 如下,见代码: struct TreeNode { int va…