[leetcode] permutations 排列

写了两个，一个是直接的递归实现：

class Solution {

public:



    void swap(vector<int> &num,int left,int right)

    {

        num[left] = num[left]^num[right];

        num[right] = num[left]^num[right];

        num[left] = num[left]^num[right];

    }



    void permuteHelp(vector<int> &num,int fix,vector< vector<int> > &result)

    {

        if(fix==num.size()-1)

        {

            result.push_back(num);

            return;

        }

        permuteHelp(num,fix+1,result);

        for(int i=fix+1;i<num.size();i++)

        {

            swap(num,i,fix);

            permuteHelp(num,fix+1,result);

            swap(num,i,fix);

        }

    }



    vector<vector<int> > permute(vector<int> &num) {

        vector< vector<int> > result;

        if (num.size()<1)

            return result;

        if(num.size()<2)

            result.push_back(num);

        else

            permuteHelp(num,0,result);

        return result;

    }

};

 

另外一个是通过类似STL 的 next_permutation 函数方法实现：

下面链接是 STL 的 next_permutation 函数 描述：

http://www.cnblogs.com/Azhu/articles/3897586.html

按字典左起从小到达顺序给出当前序列的下一个排序，适用于序列中有重复元素的情况，函数的过程是：

1.右起寻找相邻的两数，满足next<next1

2.右起寻找第一个大于next的数，记为mid

3.交换 mext 与mid

4.逆序next1 （包括next1）到右末尾

class Solution{

public:



    int factorial(int n)

    {

        return n<2?1:factorial(n-1)*n;

    }



    void next_per(vector<int> & num)

    {

        vector<int>::iterator first,last,next,next1,mid;

        first = num.begin();

        last = num.end();

        next = last;

        if(first==--next||first==last)

            return ;

        while(1)

        {

            next1=next--;

            if(*next<*next1)

            {

                mid = last;

                while(!(*next<*--mid));

                iter_swap(next,mid);

                reverse(next1,last);

                return ;

            }

            if(next==first)

            {

                reverse(first,last);

                return ;

            }

        }

    }



    vector< vector<int> > permute(vector<int> &num)

    {

        vector< vector<int> > result;

        if(num.size()<1)

            return result;

        sort(num.begin(),num.end());

        result.push_back(num);

        for(int i=1;i<factorial(num.size());i++)

        {

            next_per(num);

            result.push_back(num);

        }

        return result;

    }

};

 

main函数：

#include <vector>

#include <iostream>

#include <algorithm>

using namespace std;

int main()

{

    vector<int> num;

    int a = 2;

    for(int i=0;i<a;i++)

        num.push_back(i);

    Solution solution;

    vector< vector<int> > result;

    result = solution.permute(num);

    for(int id = 0;id<result.size();id++)

    {

        for(int i=0;i<a;i++)

            cout<<result[id][i]<<' ';

        cout<<endl;

    }

    cout<<result.size()<<endl;

    return 0;

}