标签：return 重复 otto pos start 个数 eve false code
问题：
给一组数组，对其进行一个排序得到新的数组res，使得对res做一下操作最后得到target为一个递增数组。
1.取第一个数，到target数组。
2.把剩下的第一个数，排到原数组res最后。
3.重复1和2，直到原数组res为空。
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 <= A.length <= 1000 1 <= A[i] <= 10^6 A[i] != A[j] for all i != j
解法：
要求res数组，那么我们把res的各位index取出。
(n=deck.size())
即为：0,1,2,...,n
做题目的操作，最后得到递增的target，
那么最后经过操作得到的pos序列(index的一个排序)，对应上sort后的target每一位。
即target[i]应该在res的pos[i]位上。
再将target的每一位放入相应的index上。也即是恢复了res数组。
res[pos[i]]=target[i]
代码参考：
1 class Solution { 2 public: 3 vector<int> deckRevealedIncreasing(vector<int>& deck) { 4 int n=deck.size(); 5 vector<int> pos; 6 vector<int> res(n,0); 7 queue<int> q; 8 sort(deck.begin(), deck.end()); 9 for(int i=0; i<n; i++){ 10 q.push(i); 11 } 12 pos.push_back(q.front()); 13 q.pop(); 14 while(!q.empty()){ 15 q.push(q.front()); 16 q.pop(); 17 pos.push_back(q.front()); 18 q.pop(); 19 } 20 for(int i=0; i<n; i++){ 21 res[pos[i]]=deck[i]; 22 } 23 return res; 24 } 25 };
950. Reveal Cards In Increasing Order
标签：return 重复 otto pos start 个数 eve false code
原文地址：https://www.cnblogs.com/habibah-chang/p/12971474.html
