<p>Given an array of integers <code>nums</code>, sort the array in ascending order and return it.</p>

<p>You must solve the problem <strong>without using any built-in</strong> functions in <code>O(nlog(n))</code> time complexity and with the smallest space complexity possible.</p>

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

<pre>
<strong>Input:</strong> nums = [5,2,3,1]
<strong>Output:</strong> [1,2,3,5]
<strong>Explanation:</strong> After sorting the array, the positions of some numbers are not changed (for example, 2 and 3), while the positions of other numbers are changed (for example, 1 and 5).
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,1,1,2,0,0]
<strong>Output:</strong> [0,0,1,1,2,5]
<strong>Explanation:</strong> Note that the values of nums are not necessairly unique.
</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>-5 * 10<sup>4</sup> &lt;= nums[i] &lt;= 5 * 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    vector<int> sortArray(vector<int> nums) {
        // recursive top down mergesort
        // note that you deleted the reference type to this function argument
        if (nums.size() < 2) {
            return nums;
        }
        int size = nums.size();
        vector<int> n1 = sortArray(vector<int> (nums.begin(), nums.begin() + size/2));
        vector<int> n2 = sortArray(vector<int> (nums.begin() + size/2, nums.end()));
        return merge(n1, n2);
    }
private:
    vector<int> merge(vector<int>& n1, vector<int>& n2) {
        vector<int> res (n1.size() + n2.size());
        int p1 = 0;
        int p2 = 0;
        int ptr = 0;
        while (p1 < n1.size() && p2 < n2.size()) {
            if (n1[p1] <= n2[p2]) {
                res[ptr++] = n1[p1++];
            } else {
                res[ptr++] = n2[p2++];
            }
        }
        while (p1 < n1.size()) {
            res[ptr++] = n1[p1++];
        }
        while (p2 < n2.size()) {
            res[ptr++] = n2[p2++];
        }
        return res;
    }
};
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