<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) {
   std::vector<int> sortedNums = nums; // Copy original numbers  
    std::sort(sortedNums.begin(), sortedNums.end()); // Sort the copy  
    
    int n = nums.size();  
    std::vector<int> result(n);  
    
    // The middle index for wiggle sorting  
    int mid = (n - 1) / 2; // mid will point to the last item of the left half  
    int j = n - 1; // Last element index of sorted array  

    // Fill result array using the two halves of the sorted array  
    for (int i = 0; i < n; ++i) {  
        if (i % 2 == 0) {  
            result[i] = sortedNums[mid--]; // Fill even indices with high-value elements  
        } else {  
            result[i] = sortedNums[j--]; // Fill odd indices with low-value elements  
        }  
    }  

    nums = result; // Update original vector with the new order  

}  
};
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