[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;
}