<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();
        sort(nums.begin(), nums.end());
        if (n == 1 || n == 2) {
            return;
        }

        vector<int> front;
        vector<int> rear;
        int ptr1 = 0;
        int ptr2 = n - 1;
        while (ptr1 < ptr2) {
            front.push_back(nums[ptr1]);
            rear.push_back(nums[ptr2]);
            ptr1++;
            ptr2--;
        }
        if (n % 2 != 0) {
            front.push_back(nums[ptr1]);
        }

        int ptrf = front.size() - 1;
        for (int i = 0; i < n; i += 2) {
            nums[i] = front[ptrf];
            ptrf--;
        }

        int ptrr = 0;
        for (int j = 1; j < n; j += 2) {
            nums[j] = rear[ptrr];
            ptrr++;
        }

    }
};
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