<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) {
        if(nums.size() == 1)
            return;
        if(nums.size() == 2){
            if(nums[0] > nums[1]){
                int t = nums[0];
                nums[0] = nums[1];
                nums[1] = t;
            }
            return;
        }
        sort(nums.begin(), nums.end());
        vector<int> v;
        int i=0, j = nums.size() - 1;
        while(i <= j){
            v.push_back(nums[i]);
            v.push_back(nums[j]);
            i++; j--;
        }

        j = v.size() - 1; 
        int z = 0;
        while(j > 0){
            if(v[j] == v[j-1]){
                int t = v[j-1];
                v[j-1] = v[z];
                v[z] = t;
                z += 2;
            }
            j -= 2;
        }
        for(int k=0;k<nums.size();k++)
            nums[k] = v[k];
    }
};
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