Permutation

问题描述：

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].

 

C++ STL中提供了next_permutation()函数，在OJ中可以很容易的实现，下面直接实现了STL 中next_permutation()函数

#include <vector>
#include <algorithm>  //next_permutaion,iter_swap,reverse 
using std::vector;

namespace ns {
    template <class BidirectionalIterator>
    bool next_permutation (BidirectionalIterator first,
                         BidirectionalIterator last)
    {
        if(first == last)  return false;  //empty

        BidirectionalIterator i = last;
        if((--i)==first)   return false;  //just one element

        while(i != first)
        {
            auto t = i;
            --i;
            if(*i < *t)
            {
                auto j = last;
                while(!(*--j > *i));
                iter_swap(i,j);
                reverse(t, last);
                return true;
            }
        }
    reverse(first, last);
    return false;
    }
} 

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

int main()
{
    vector<int> v{1,2,3};
    vector<vector<int>> res;
    Solution s;
    res = s.permute(v);

    return 0;
}

 方法二：递归实现

#include <vector>
#include <algorithm>   
#include <iostream>
using std::vector;

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        vector<int> v;
        dfs(nums,v,res);
        return res;
    }
private:
    void dfs(vector<int>& nums,vector<int>& v,vector<vector<int>>& res)
    {
        if(v.size() == nums.size())
        {
            res.push_back(v);
            return;
        }
        for(auto i : nums)
        {
            auto it = find(v.begin(),v.end(),i);
            if(it == v.end())
            {
                v.push_back(i);
                dfs(nums,v,res);
                v.pop_back();
            }
        }
    }
};

int main()
{
    vector<int> v{1,2,3};
    vector<vector<int>> res;
    Solution s;
    res = s.permute(v);
    for (int i = 0; i < res.size(); ++i)
    {
        for(int j=0;j<res[i].size();j++)
        {
            cout << res[i][j];
        }
        cout << endl;
    }
    return 0;
}