<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 quickSort(vector<int>& N, int l, int r) {
        if (l >= r) return;
        
        swap(N[l], N[l + rand() % (r - l)]);
        int p = l++, e = r;
        while (l <= r) {
            if (N[l] > N[p] && N[r] < N[p]) {
                swap(N[l++], N[r--]);
            }
            if (N[l] <= N[p]) l++;
            if (N[r] >= N[p]) r--;
        }
        swap(N[p], N[r]);
        
        quickSort(N, p, r - 1);
        quickSort(N, r + 1, e);
    }
    
    void mergeSort(vector<int>& N, int l, int r) {
        if (l >= r) return;
        if (r - l == 1 && N[l] > N[r]) swap(N[l], N[r]);
        
        int m = l + (r - l) / 2;
        mergeSort(N, l, m);
        mergeSort(N, m + 1, r);
        
        vector<int> tmp;
        int ll = l, rr = m + 1;
        while (ll <= m || rr <= r) {
            if (ll > m) {
                tmp.push_back(N[rr++]);
            } else if (rr > r) {
                tmp.push_back(N[ll++]);
            } else {
                tmp.push_back((N[rr] < N[ll])? N[rr++]: N[ll++]);
            }
        }
        
        for (int i = 0; i < tmp.size(); ++i) {
            N[i+l] = tmp[i];
        }
    }
    
    vector<int> sortArray(vector<int>& N) {
        mergeSort(N, 0, N.size() - 1);
        
        return N;
    }
};
265028