LeetCode
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47. Permutations II Problem's Link
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;
}
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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