<p>Given an integer array <code>nums</code>, reorder it such that <code>nums[0] &lt; nums[1] &gt; nums[2] &lt; nums[3]...</code>.</p>

<p>You may assume the input array always has a valid answer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5,1,1,6,4]
<strong>Output:</strong> [1,6,1,5,1,4]
<strong>Explanation:</strong> [1,4,1,5,1,6] is also accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,2,2,3,1]
<strong>Output:</strong> [2,3,1,3,1,2]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 5000</code></li>
	<li>It is guaranteed that there will be an answer for the given input <code>nums</code>.</li>
</ul>

<p>&nbsp;</p>
<strong>Follow Up:</strong> Can you do it in <code>O(n)</code> time and/or <strong>in-place</strong> with <code>O(1)</code> extra space?
class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        int left = 0, right = nums.size() - 1;
        while (true) {
            int pivot = rand() % (right - left + 1) + left;
            int newpivot = partion(nums, left, right, pivot);
            if (newpivot == nums.size() - k)
                return nums[newpivot];
            else if (newpivot > nums.size() - k)
                right = newpivot - 1;
            else
                left = newpivot + 1;
        }
    }
    int partion(vector<int>& nums, int left, int right, int pivot) {
        int pivotval = nums[pivot];
        swap(nums[right], nums[pivot]);
        int storedval = left;
        for (int i = left; i < right; i++) {
            if (nums[i] < pivotval) {
                swap(nums[i], nums[storedval]);
                storedval++;
            }
        }
        swap(nums[right], nums[storedval]);
        return storedval;
    }

    void wiggleSort(vector<int>& nums) {
        int median = findKthLargest(nums, (nums.size() + 1) / 2);
        int n = nums.size();

        int left = 0, i = 0, right = n - 1;

        while (i <= right) {

            if (nums[newIndex(i, n)] > median) {
                swap(nums[newIndex(left++, n)],nums[ newIndex(i++, n)]);
            } else if (nums[newIndex(i, n)] < median) {
                swap(nums[newIndex(right--, n)], nums[newIndex(i, n)]);
            } else {
                i++;
            }
        }
    }

    int newIndex(int index, int n) { return (1 + 2 * index) % (n | 1); }
};
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