2019-03-27 14:10:37 问题描述: 问题求解: 模拟题.考虑角度是从结果来进行反推. input - [2,3,5,7,11,13,17] (just sort the input that you get) The last number that you wanna get is the last number in the array - (17) The penultimate number is 13. So put 13 on top of 17 (13,17) an…

In a deck of cards, every card has a unique integer.  You can order the deck in any order you want. Initially, all the cards start face down (unrevealed) in one deck. Now, you do the following steps repeatedly, until all cards are revealed: Take the…

Difficulty: Medium Problem In a deck of cards, every card has a unique integer. You can order the deck in any order you want. Initially, all the cards start face down (unrevealed) in one deck. Now, you do the following steps repeatedly, until all car…

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 模拟 日期 题目地址:https://leetcode.com/problems/reveal-cards-in-increasing-order/description/ 题目描述 In a deck of cards, every card has a unique integer. You can order the deck in any o…

In a deck of cards, every card has a unique integer.  You can order the deck in any order you want. Initially, all the cards start face down (unrevealed) in one deck. Now, you do the following steps repeatedly, until all cards are revealed: Take the…

In a deck of cards, every card has a unique integer.  You can order the deck in any order you want. Initially, all the cards start face down (unrevealed) in one deck. Now, you do the following steps repeatedly, until all cards are revealed: Take the…

题目如下: In a deck of cards, every card has a unique integer.  You can order the deck in any order you want. Initially, all the cards start face down (unrevealed) in one deck. Now, you do the following steps repeatedly, until all cards are revealed: Tak…

牌组中的每张卡牌都对应有一个唯一的整数.你可以按你想要的顺序对这套卡片进行排序. 最初,这些卡牌在牌组里是正面朝下的(即,未显示状态). 现在,重复执行以下步骤,直到显示所有卡牌为止: 从牌组顶部抽一张牌,显示它,然后将其从牌组中移出. 如果牌组中仍有牌,则将下一张处于牌组顶部的牌放在牌组的底部. 如果仍有未显示的牌,那么返回步骤 1.否则,停止行动. 返回能以递增顺序显示卡牌的牌组顺序. 答案中的第一张牌被认为处于牌堆顶部. 示例: 输入:[17,13,11,2,3,5,7] 输出:[2,13…

#include<iostream> using namespace std; int main() { int n,i,j,m,k; cin>>n; int a[n];  for(i=0;i<n;i++){ cin>>a[i]; } for(i=0;i<(n-1);i++){ k=i; for(j=i+1;j<n;j++){ if(a[k]>a[j]){ m=a[i],a[i]=a[j],a[j]=m; } } } for(i=0;i<n…

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