950. Reveal Cards In Increasing Order

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：

class Solution {
public:
    vector<int> deckRevealedIncreasing(vector<int>& deck) {
        int n=deck.size();
        vector<int> pos;
        vector<int> res(n,0);
        queue<int> q;
        sort(deck.begin(), deck.end());
        for(int i=0; i<n; i++){
            q.push(i);
        }
        pos.push_back(q.front());
        q.pop();
        while(!q.empty()){
            q.push(q.front());
            q.pop();
            pos.push_back(q.front());
            q.pop();
        }
        for(int i=0; i<n; i++){
            res[pos[i]]=deck[i];
        }
        return res;
    }
};