<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:
    void wiggleSort(vector<int>& nums) {
        priority_queue<int> pq(nums.begin(),nums.end());
        int n = nums.size();
        int x = n/2;
        vector<int> mx(x);
        for(int i=0;i<x;i++){
            int a = pq.top();;
            mx[i] = a;
            pq.pop();
        }
        vector<int> mn(n-x);
        for(int i=0;i<n-x;i++){
            int a = pq.top();;
            mn[i] = a;
            pq.pop();
        }
        int i = 0;
        int j = 0;
        int k = 0;
        while(i<x and j<n-x){
            if(k%2!=0){
               nums[k++] = mx[i++]; 
            }
            else nums[k++] = mn[j++]; 
        }
        while(i<x){
            nums[k++] = mx[i++]; 
        }
        while(j<n-x){
            nums[k++] = mn[j++]; 
        }
    }
};
21900