leetcode 46. Permutations

Given a collection of distinct integers, return all possible permutations.

Example:

Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

思路一：二话不说，经典题， 回溯。

 1 class Solution {
 2 public:
 3     vector<vector<int>> permute(vector<int>& nums) {
 4         sort(nums.begin(), nums.end());
 5         int len = nums.size();
 6         vector<bool> visited(len, false);
 7         vector<vector<int> > res;
 8         vector<int> tmp;
 9         dfs(0, len, nums, tmp, res, visited);
10         return res;
11     }
12 private:
13     void dfs(int start, int len, vector<int> &nums, vector<int> &v, vector<vector<int> > &res, vector<bool> &visited) {
14         if (start == len) {
15             res.push_back(v);
16             return ;
17         }
18         for (int i = 0; i < len; i++) {
19             if (visited[i] == false) {
20                 visited[i] = true;
21                 v.push_back(nums[i]);
22                 dfs(start + 1, len, nums, v, res, visited);
23                 v.pop_back();
24                 visited[i] = false;
25             }
26         }
27     }
28 };

思路二：

 1 class Solution {
 2 public:
 3     vector<vector<int>> permute(vector<int>& nums) {
 4         sort(nums.begin(), nums.end());
 5         int len = nums.size();
 6         vector<vector<int> > res;
 7         dfs(0, len, nums, res);
 8         return res;
 9     }
10 private:
11     void dfs(int start, int len, vector<int> &nums, vector<vector<int> > &res) {
12         if (start == len) {
13             res.push_back(nums);
14             return ;
15         }
16         for (int i = start; i < len; i++) {
17             swap(nums[start], nums[i]);
18             dfs(start + 1, len, nums, res);
19             swap(nums[start], nums[i]);
20         }
21     }
22 };

 

posted @ 2019-09-18 16:35  琴影  阅读(197)  评论(0)  编辑  收藏  举报