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?

含义：给定一个没有排序的数组，重排以后达到 nums[0] < nums[1] > nums[2] < nums[3]....这种效果

    public void wiggleSort(int[] nums) {
        //        先给数组排序，然后找到数组的中间的数，相当于把有序数组从中间分成两部分，然后从前半段的末尾取一个，在从后半的末尾去一个，以此类推直至都取完
        int[] temp = (int[])nums.clone();
        Arrays.sort(temp);
        int n = nums.length,k=(1+n)/2,j=n;
        boolean flag = true;
        for (int i=0;i<n;i++)
        {
            nums[i] = flag?temp[--k]:temp[--j];
            flag = !flag;
        }        
    }