324. Wiggle Sort II

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

Example 1:

Input: nums = [1, 5, 1, 1, 6, 4]
Output: One possible answer is [1, 4, 1, 5, 1, 6].

Example 2:

Input: nums = [1, 3, 2, 2, 3, 1]
Output: 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?

解题思路：

我们可以将首先把nums排序，然后从中间和尾部分别往前向新数组添加数字

注意的是：

　　当nums有奇数个数字时，选取的中点为 n/2

　　当nums有偶数个数字时，选取的终点为n/2 - 1

此时时间复杂度为O(nlogn) 因为用了排序方法

空间复杂度为O(n)

代码：

class Solution {
public:
    void wiggleSort(vector<int>& nums) {
        vector<int> temp(nums.begin(), nums.end());
        sort(temp.begin(), temp.end());
        int n = nums.size();
        int m = n&1 ? n/2 : n/ 2 - 1 ;
        int j = n - 1;
        for(int i = 0; i < n; i++){
            nums[i] = (i & 1) ? temp[j--] : temp[m--];
        }
    }
};

Time O(n) & Space O(1)解法：

https://leetcode.com/problems/wiggle-sort-ii/discuss/77681/O(n)-time-O(1)-space-solution-with-detail-explanations