LeetCode46,47 Permutations, Permutations II

题目：

LeetCode46 I

Given a collection of distinct numbers, return all possible permutations. （Medium）

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

[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

LeetCode47 II

Given a collection of numbers that might contain duplicates, return all possible unique permutations.（Medium）

For example,
[1,1,2] have the following unique permutations:

[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

分析： 首先先用next_permutation解这两个题。用好STL。

Permutations I：

 1 class Solution {
 2 public:
 3     vector<vector<int>> permute(vector<int>& nums) {
 4         vector<vector<int>> result;
 5         int len = nums.size();
 6         sort(nums.begin(), nums.end());
 7         do {
 8             result.push_back(nums);
 9         } while (next_permutation(nums.begin(),nums.end()));
10
11         return result;
12     }
13 };

Permutations II：

 1 class Solution {
 2 public:
 3     vector<vector<int>> permuteUnique(vector<int>& nums) {
 4         vector<vector<int>> result;
 5         int len = nums.size();
 6         sort(nums.begin(), nums.end());
 7         do {
 8             result.push_back(nums);
 9         } while (next_permutation(nums.begin(),nums.end()));
10
11         return result;
12     }
13 };

当然，用STL写好的函数肯定不是面试的目的，permutation的递归方法也是经典的回溯算法题。

代码：

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