LeetCode - 47. Permutations II

简介: 47. Permutations II  Problem's Link  ---------------------------------------------------------------------------- Mean:  给定一个数组(元素可能重复),求这个数组的全排列.

47. Permutations II 

Problem's Link

 ----------------------------------------------------------------------------

Mean: 

给定一个数组(元素可能重复),求这个数组的全排列.

analyse:

注意需要先排序,和上一题的区别在于当发现要交换的两数相等时,无需再往下递归,避免了重复的排列.

Time complexity: O(N)

 

view code

#include <bits/stdc++.h>
using namespace std;

class Solution
{
public :
    vector < vector < int >> permuteUnique( vector < int > nums)
    {
        sort( nums . begin (), nums . end());
        vector < vector < int >> res;
        permutate( res , nums , 0);
        return res;
    }

    void permutate( vector < vector < int >>& res , vector < int > nums , int begin)
    {
        if( begin >= nums . size())
            res . push_back( nums);
        for( int i = begin; i < nums . size(); ++ i)
        {
            if( i != begin && nums [ i ] == nums [ begin ])
                continue;
            swap( nums [ i ], nums [ begin ]);
            permutate( res , nums , begin + 1);
        }
    }
};

int main()
{
    int n;
    while( cin >>n)
    {
        vector < int > nums(n);
        for( int i = 0; i <n; ++ i)
            cin >> nums [ i ];
        Solution solution;
        auto ans = solution . permuteUnique( nums);
        for( auto p1: ans)
        {
            for( auto p2: p1)
            {
                cout << p2 << " ";
            }
            cout << endl;
        }
    }
    return 0;
}
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