Next Permutation [Leetcode]

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
Solution: find out the transform pattern first. The basic idea is: scan the array from the end to the beginning to find out the first "less than" pair <num[i], num[i+1>, swap the num[i] in the pair with the the smallest element but greater than num[i] in the subarray of num[i+1, ..., n-1], where n is the size of the array num. Then, we need to reverse the subarrary num[n+1, ..., n-1]. If no such "less than" pair is found, then we can easily reverse the entire array num to get the result.  [Think carefully first, and then program....]

class Solution {
public:
    void nextPermutation(vector<int> &num) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int n = num.size();
        if(n!=0){
            for(int i=n-2; i>=0; --i){
                if(num[i]<num[i+1]){
                    int j=0, k=0;
                    //find the smallest one but larger than num[i] in num[i+1] to num[n-1]
                    for(j=n-1; j>i; j--)
                        if(num[j]>num[i]){
                            swap(num[i], num[j]);
                            break;
                        }
                    //reverse num[i+1] to num[n-1]
                    reverse(num.begin()+i+1, num.end());
                    return;
                }
            }
            reverse(num.begin(), num.end());
        }
    }
};