113th LeetCode Weekly Contest Reveal Cards In Increasing Order

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:

  1. Take the top card of the deck, reveal it, and take it out of the deck.
  2. If there are still cards in the deck, put the next top card of the deck at the bottom of the deck.
  3. If there are still unrevealed cards, go back to step 1.  Otherwise, stop.

Return an ordering of the deck that would reveal the cards in increasing order.

The first entry in the answer is considered to be the top of the deck.

Example 1:

Input: [17,13,11,2,3,5,7]
Output: [2,13,3,11,5,17,7]
Explanation:
We get the deck in the order [17,13,11,2,3,5,7] (this order doesn‘t matter), and reorder it.
After reordering, the deck starts as [2,13,3,11,5,17,7], where 2 is the top of the deck.
We reveal 2, and move 13 to the bottom.  The deck is now [3,11,5,17,7,13].
We reveal 3, and move 11 to the bottom.  The deck is now [5,17,7,13,11].
We reveal 5, and move 17 to the bottom.  The deck is now [7,13,11,17].
We reveal 7, and move 13 to the bottom.  The deck is now [11,17,13].
We reveal 11, and move 17 to the bottom.  The deck is now [13,17].
We reveal 13, and move 17 to the bottom.  The deck is now [17].
We reveal 17.
Since all the cards revealed are in increasing order, the answer is correct.

Note:

  1. 1 <= A.length <= 1000
  2. 1 <= A[i] <= 10^6
  3. A[i] != A[j] for all i != j

倒过来看就能找到规律

class Solution {
public:
    vector<int> deckRevealedIncreasing(vector<int>& deck) {
        int s = deck.size();
        sort(deck.begin(),deck.end());
        if(s==1){
            return deck;
        }
        if(s==2){
            return deck;
        }
        vector<int> result;
        result.push_back(deck[s-2]);
        result.push_back(deck[s-1]);
        for(int i=s-3;i>=0;i--){

            int len = result.size();
            int last= result[len-1];
            result.insert(result.begin(),last);

            result.pop_back();
            result.insert(result.begin(),deck[i]);

        }
        return result;
    }
};

原文地址:https://www.cnblogs.com/yinghualuowu/p/10055366.html

时间: 2024-10-09 02:37:22

113th LeetCode Weekly Contest Reveal Cards In Increasing Order的相关文章

[Swift Weekly Contest 113]LeetCode950. 按递增顺序显示卡牌 | Reveal Cards In Increasing Order

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

113th LeetCode Weekly Contest Flip Equivalent Binary Trees

For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees. A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip opera

113th LeetCode Weekly Contest Largest Time for Given Digits

Given an array of 4 digits, return the largest 24 hour time that can be made. The smallest 24 hour time is 00:00, and the largest is 23:59.  Starting from 00:00, a time is larger if more time has elapsed since midnight. Return the answer as a string

Leetcode Weekly Contest 86

Weekly Contest 86 A:840. 矩阵中的幻方 3 x 3 的幻方是一个填充有从 1 到 9 的不同数字的 3 x 3 矩阵,其中每行,每列以及两条对角线上的各数之和都相等. 给定一个由整数组成的 N × N 矩阵,其中有多少个 3 × 3 的 "幻方" 子矩阵?(每个子矩阵都是连续的). 直接模拟即可,本来是签到题,由于粗心,浪费了时间. 1 class Solution { 2 public: 3 int numMagicSquaresInside(vector&l

Leetcode Weekly Contest 152

退役老人现在连leetcode都不会做了 = = 今天早上做了leetcode第三题题目看错了,加上比赛中间还在调投稿的实验,一心二用直接gg 总结下教训就是 本渣现在做题连题目都看不清就开始做.开始写题之前应当把样例过一遍,然后自己再造1-2个例子,然后再开始做 A题:统计素数的个数(素数筛或者sqrt(n)判断都可以),然后分别计算count! class Solution { public: int numPrimeArrangements(int n) { vector<int> ha

[Leetcode Weekly Contest]174

链接:LeetCode [Leetcode]5328. 方阵中战斗力最弱的 K 行 给你一个大小为?m?* n?的方阵?mat,方阵由若干军人和平民组成,分别用 0 和 1 表示. 请你返回方阵中战斗力最弱的?k?行的索引,按从最弱到最强排序. 如果第?i?行的军人数量少于第?j?行,或者两行军人数量相同但 i 小于 j,那么我们认为第 i 行的战斗力比第 j 行弱. 军人 总是 排在一行中的靠前位置,也就是说 1 总是出现在 0 之前. 输入:\(mat = [[1,1,0,0,0], [1,

108th LeetCode Weekly Contest Binary Subarrays With Sum

In an array A of 0s and 1s, how many non-empty subarrays have sum S? Example 1: Input: A = [1,0,1,0,1], S = 2 Output: 4 Explanation: The 4 subarrays are bolded below: [1,0,1,0,1] [1,0,1,0,1] [1,0,1,0,1] [1,0,1,0,1] Note: A.length <= 30000 0 <= S <

108th LeetCode Weekly Contest Minimum Falling Path Sum

Given a square array of integers A, we want the minimum sum of a falling path through A. A falling path starts at any element in the first row, and chooses one element from each row.  The next row's choice must be in a column that is different from t

116th LeetCode Weekly Contest Maximum Width Ramp

Given an array A of integers, a ramp is a tuple (i, j) for which i < j and A[i] <= A[j].  The width of such a ramp is j - i. Find the maximum width of a ramp in A.  If one doesn't exist, return 0. Example 1: Input: [6,0,8,2,1,5] Output: 4 Explanatio