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