<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 _print(vector<int>& vec) {
        for (int i = 0; i < vec.size(); ++i) {
            cout << vec[i] << " ";
        }
        cout << endl;
    }
    vector<int> _sort(vector<int>& nums, int start, int end) {
        if (end - start < 2) {
            return {nums[start]};
        }
        if (end - start == 2) {
            if (nums[start] > nums[end-1]) {
                return {nums[end-1], nums[start]};
            }
            return {nums[start], nums[end-1]};
        }
        int mid = (end - start)/2 + start;
        vector<int> first = _sort(nums, start, mid);
        vector<int> second = _sort(nums, mid, end);
        vector<int> v(end-start);
        int i = 0, j = 0;
        while (i <= first.size() || j <= second.size()) {
            if (i == first.size() && j == second.size()) break;
            if (i == first.size()) {
                v[i+j] = second[j];
                j++;
                continue;
            }
            if (j == second.size()) {
                v[i+j] = first[i];
                i++;
                continue;
            }
            if (first[i] < second[j]) {
                v[i+j] = first[i];
                i++;
                continue;
            }
            if (first[i] > second[j]) {
                v[i+j] = second[j];
                j++;
                continue;
            }
            v[i+j] = first[i];
            i++;
        }
        return v;
    }
public:
    vector<int> sortArray(vector<int>& nums) {
        return _sort(nums, 0, nums.size());
    }
};

// split
// swap
// move pointer merge
//
160853