47. Permutations II
Problem's Link
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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;
}
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;
}