<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) {
        int n=nums.size();
        nth_element(nums.begin(),nums.begin()+(n/2),nums.end());
        int i=0,j=0,m=nums[n/2];
        n--;
        while(j<=n)
        {
            if(nums[j]<m)
                swap(nums[i++],nums[j++]);
            else if(nums[j]>m)
                swap(nums[j],nums[n--]);
            else
                j++;
        }
        n=nums.size();
        vector<int> res(n);
        j=(n%2!=0)?(n/2):(n/2)-1;
        for(int i=0;i<n;i+=2)
        {
            res[i]=nums[j--];
        }
        j=n-1;
        for(int i=1;i<n;i+=2)
        {
            res[i]=nums[j--];
        }
        nums=res;
    }
};
20400