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：

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> result;
        int len = nums.size();
        sort(nums.begin(), nums.end());
        do {
            result.push_back(nums);
        } while (next_permutation(nums.begin(),nums.end()));
    
        return result;
    }
};

Permutations II：

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> result;
        int len = nums.size();
        sort(nums.begin(), nums.end());
        do {
            result.push_back(nums);
        } while (next_permutation(nums.begin(),nums.end()));
    
        return result;
    }
};

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：

class Solution {
private:
    vector<vector<int>> result;
    void helper(vector<int>& nums, int start, int end) {
        if (start == end) {
            result.push_back(nums);
        }
        for (int i = start; i <= end; ++i) {
            swap(nums[start], nums[i]);
            helper(nums, start + 1, end);
            swap(nums[start], nums[i]);
        }
    }
public:
    vector<vector<int>> permute(vector<int>& nums) {
        helper(nums, 0, nums.size() - 1);
        return result;
    }
};