324. Wiggle Sort II

Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3]....

Example:

(1) Given nums = [1, 5, 1, 1, 6, 4], one possible answer is [1, 4, 1, 5, 1, 6]. (2) Given nums = [1, 3, 2, 2, 3, 1], one possible answer is [2, 3, 1, 3, 1, 2].

Note:
You may assume all input has valid answer.

Follow Up:
Can you do it in O(n) time and/or in-place with O(1) extra space?

https://discuss.leetcode.com/topic/32929/o-n-o-1-after-median-virtual-indexing
思路：先找到中位数，然后从左往右将比中位数大的数放在奇数位置上，从右往左将比中位数小的数放在偶数位置上．

其中(1+2*i)%(n|1)这是一个设计极其巧妙的映射，这样会使得小于一半的数映射到1, 3, 5等奇数位置，也就是low会映射到的奇数位置，而大于一半的数会映射到从高到低的偶数位置，也就是high会依次映射到从(n|1)到０的偶数位置．这样就会让小于中位数的值优先从右往左放，而大于中位数的值优先从左往右放，剩下的等于中位数数的值会留在大于中位数和小于中位数的值中间．

Accessing A(0) actually accesses nums[1].Accessing A(1) actually accesses nums[3].Accessing A(2) actually accesses nums[5].Accessing A(3) actually accesses nums[7].Accessing A(4) actually accesses nums[9].Accessing A(5) actually accesses nums[0].Accessing A(6) actually accesses nums[2].Accessing A(7) actually accesses nums[4].Accessing A(8) actually accesses nums[6].Accessing A(9) actually accesses nums[8].

class Solution {public:    void wiggleSort(vector<int>& nums) {        int n = nums.size();        nth_element(nums.begin(),nums.begin()+n/2,nums.end());        int mid = nums[n/2];        int i =0 , low =0, high = n-1;        #define A(i) nums[(1+2*(i)) % (n | 1)]        while(i <= high){            if(A(i) > mid){                swap(A(i++),A(low++));            }            else if(A(i) < mid){                swap(A(i),A(high--));            }            else i++;        }    }};