<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> *arr;
    vector<int> sortArray(vector<int>& nums) {
        arr = &nums;
        mergeSort(0, nums.size()-1);
        return nums;
    }
    void mergeSort(int l, int r){
        if(l>=r) return;
        int m = (l+r)/2;
        mergeSort(l, m);
        mergeSort(m+1, r);
        merge(l, m, r);
    }
    void merge(int l, int m, int r){
        vector<int> tmp;
        int i=l, j=m+1;
        while(i<=m || j<=r){
            if(j>r || (i<=m && (*arr)[i]<=(*arr)[j])){
                tmp.push_back((*arr)[i++]);
            }else{
                tmp.push_back((*arr)[j++]);
            }
        }
        for(int k=l; k<=r; k++) (*arr)[k]=tmp[k-l];
    }
};
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