[Leetcode] Permutations

Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

与Next Permutation一样，n个元素的排列共有n！个，所以只要执行n！次next_permutation就行。

 1 class Solution {
 2 public:
 3     bool nextPermutation(vector<int> &num) {
 4         if (num.size() < 2) {
 5             return false;
 6         }
 7         int i, k;
 8         bool flag = true;
 9         for (i = num.size() - 2; i >= 0; --i) {
10             if (num[i] < num[i + 1]) {
11                 break;
12             }
13         }
14         if (i < 0) {
15             flag = false;
16         }
17         for (k = num.size() - 1; k > i; --k) {
18             if (num[k] > num[i]) {
19                 break;
20             }
21         }
22         swap(num[i], num[k]);
23         reverse(num.begin() + i + 1, num.end());
24         return flag;
25     }
26     
27     long long getNum(int n) {
28         long long res = 1;
29         while (n > 0) {
30             res *= n;
31             n--;
32         }
33         return res;
34     }
35     
36     vector<vector<int> > permute(vector<int> &num) {
37         vector<vector<int> > res;
38         res.push_back(num);
39         long long n = getNum(num.size());
40         while (--n) {
41             nextPermutation(num);
42             res.push_back(num);
43         }
44         return res;
45     }
46 };