第一遍想太多. 其实做一个fall-through就能把问题解决，原因是因为：这样的fall-through每次在乎两个element，可以一口气搞定，无关乎再之前的elements。
特别的一点：flag来巧妙的掌控山峰和低谷的变化。又是神奇的一幕啊！

这样子的奇观，见过就要知道了，没见过的时候有点摸不着头脑。

/*
Given an unsorted array nums, reorder it in-place such that nums[0] <= nums[1] >= nums[2] <= nums[3]....

For example, given nums = [3, 5, 2, 1, 6, 4], one possible answer is [1, 6, 2, 5, 3, 4].

Tags: Array Sort
Similar Problems: (M) Sort Colors

*/

/*
Attemp2, Thoughts: (http://www.cnblogs.com/easonliu/p/4798814.html)
Draw wiggle on original array. Whenever any postion i is not working with i-1, swap them.
Cases:
1. when nums[i] is supposed to >= nums[i - 1], however it's nums[i] < nums[i - 1], swap them. For example (nums[1] vs. nums[0])
2. when nums[i] is supposed to <= nums[i - 1], however it's nums[i] > nums[i - 1], swap them. For example (nums[2] vs. nums[1])
Specially, for case 2: can times a -1 on both side, to make nums[i] * -1 < nums[i - 1] * -1. Therefore we only need 1 if statement.

Concept: whenver something does not work, fix it. (especailly now we are only taking caer of 2 elements at a time, so we can do it as a fall-through)
*/
public class Solution {
    public void wiggleSort(int[] nums) {
        if (nums == null || nums.length <= 1) {
            return;
        }
        int flag = 1;
        for (int i = 1; i < nums.length; i++) {
            if (flag * nums[i] < flag * nums[i - 1]) {
                swap(nums, i, i - 1);
            }
            flag = -1 * flag;
        }
    }

    public void swap(int[] nums, int x, int y) {
        int temp = nums[x];
        nums[x] = nums[y];
        nums[y] = temp;
    }
}


/*
Attempt1:
Wiggle sort, ofcourse can't do Arrays.sort()! Should sort in one pass.
Also, we don't need to make each wiggle elements in ascending order (1,(4),2,(5),3,(6) ... etc). I thought it over.
----
List out the example of 1,2,3,4,5,6,7,8, want to move (5,6,7,8) and insert between (1,2,3,4). It follows the follows patter:
Assume total length = n, and we are moving index i = (1 ~ n)
Step1: swap (n + i)/2, and i, (where initially i == 1)
Step2: swap (n + i)/2, and i+1
Step2: i+=2;


public class Solution {
    public void wiggleSort(int[] nums) {
        if (nums == null || nums.length == 0) {
            return;
        }
        Arrays.sort(nums);
        if (nums.length <= 2) {
            return;
        }
        int leng = nums.length;
        int ind1 = 0;
        int ind2 = 0
        for (int i = 1; i < leng; i++) {
            //Step1
            ind1 = (leng + i) / 2;
            ind2 = i;
            swap(ind1, ind2);
            //Step2:
            ind2 = i + 1;
            swap(ind1, ind2);
        }
    }

    public void swap(int[] nums, int x, int y) {
        int temp = nums[x];
        nums[x] = nums[y];
        nums[y] = temp;
    }
}
*/