<p>Given an integer array <code>nums</code> sorted in <strong>non-decreasing</strong> order, return <em>an array of <strong>the squares of each number</strong> sorted in non-decreasing order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [-4,-1,0,3,10]
<strong>Output:</strong> [0,1,9,16,100]
<strong>Explanation:</strong> After squaring, the array becomes [16,1,0,9,100].
After sorting, it becomes [0,1,9,16,100].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [-7,-3,2,3,11]
<strong>Output:</strong> [4,9,9,49,121]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code><span>1 &lt;= nums.length &lt;= </span>10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>nums</code> is sorted in <strong>non-decreasing</strong> order.</li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Squaring each element and sorting the new array is very trivial, could you find an <code>O(n)</code> solution using a different approach?
class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
    if (nums.size() == 0)
        return {};
    else if (nums.size() == 1)
    {
        std::vector<int> vec;
        vec.push_back(nums[0] * nums[0]);
        return vec;
    }

    int left = 0;
    std::stack<int> minusSt;

    while (left < nums.size() && nums[left] <= 0)
    {
        minusSt.push(-nums[left]);
        left++;
    }
    std::vector<int> res;

    while (left < nums.size() || !minusSt.empty()) {
        if (!minusSt.empty() && (left >= nums.size() || minusSt.top() < nums[left]))
        {
            res.push_back(std::pow(minusSt.top(), 2));
            minusSt.pop();
        }
        else if (left < nums.size()) {
            res.push_back(std::pow(nums[left], 2));
            left++;
        }
    }

    return res;
}
};
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