<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 {
    void merge(vector<int>& nums, int l, int m, int h){
        int p=h-l+1;

        vector<int> tmp;
        int i=l, j=m+1, k=0;

        for(int k=0; k<p; k++){
            tmp.push_back(i==m+1?nums[j++]:j==h+1?nums[i++]:nums[i]<=nums[j]?nums[i++]:nums[j++]);
        }

        i=l;
        for( k=0; k<p; ){
            nums[i++] = tmp[k++];
        }
    }
    void merge_sort(vector<int>& nums, int l, int h){
        if(l>=h) return;

        int m = l+(h-l)/2;
        merge_sort(nums, l, m);
        merge_sort(nums, m+1, h);
        merge(nums, l, m, h);
    }
public:
    vector<int> sortArray(vector<int>& nums) {
        merge_sort(nums, 0, nums.size()-1);
        return nums;
    }
};
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