# 280. Wiggle Sort

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

**Example:**

```
Input: nums = [3,5,2,1,6,4]
Output: One possible answer is [3,5,1,6,2,4]
```

对给定数组nums重排序，使其满足nums\[0] <= nums\[1] >= nums\[2] <= nums\[3]....。观察得知index奇数时比前面大, 偶数时比前面小。nums\[i]不满足时和i - 1时swap, 比如奇数时不满足则意味着nums\[i] < nums\[i - 1]，那nums\[i]定比nums\[i - 2]小, swap i 和i-1不会影响正确; 偶数时同理。

```python
class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        for i in range(1, len(nums)):
            if i & 1 == 0 and nums[i] > nums[i - 1] or i & 1 == 1 and nums[i] < nums[i - 1]:
                tmp = nums[i]
                nums[i] = nums[i - 1]
                nums[i - 1] = tmp
```


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