LeetCode 31: Next Permutation

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
The replacement must be in-place, do not allocate extra memory.
Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

public class Solution {
    public void nextPermutation(int[] num) {
        // The following algorithm generates the next permutation lexicographically after a given permutation.
        // It changes the given permutation in-place.

        // 1. Find the largest index k such that a[k] < a[k + 1]. If no such index exists, the permutation is the last permutation.
        // 2. Find the largest index l greater than k such that a[k] < a[l].
        // 3. Swap the value of a[k] with that of a[l].
        // 4. Reverse the sequence from a[k + 1] up to and including the final element a[n-1].
        int n = num.length;
        
        if (n <= 1)
            return;
        
        int k = -1;
        int l = -1;
        
        // Step 1
        for (int i = n-2; i >= 0; i--)
            if (num[i] < num[i+1])
            {
                k = i;
                break;
            }
        
        // If can't find k, num is the last permuation    
        if (k == -1)
        {
            Arrays.sort(num);
            return;
        }
        
        // Step 2
        for (int i = n-1; i > k; i--)
            if (num[i] > num[k])
            {
                l = i;
                break;
            }
            
        // Step 3
        swap(num, k, l);
        
        // Step 4
        for (int i = k+1; i < (k+1)+(n-(k+1))/2; i++)
            swap(num, i, n+(k+1)-i-1);
    }
    
    private void swap(int[] num, int i, int j)
    {
        int temp = num[i];
        num[i] = num[j];
        num[j] = temp;
    }
}