给的一个数列中，可能存在重复的数，比如 1 1 2 ，求其全排列。
记录上一个得出来的排列，看这个排列和上一个是否相同。
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;class Solution{
public:
vector<vector<int> > permuteUnique(vector<int> &num) {
vector<vector<int> > ans;
if(num.size()==0)
return ans;vector<int> _num = num;
sort(_num.begin(),_num.end());vector<int> _beforeOne = _num;
ans.push_back(_num);
while(nextPermutation(_num))
{
if(_num != _beforeOne)
{
ans.push_back(_num);
_beforeOne = _num;
}
}
return ans;
}
private:
bool nextPermutation(vector<int> &num)
{
return next_permutation(num.begin(),num.end());
}template<typename BidiIt>
bool next_permutation(BidiIt first,BidiIt last)
{
const auto rfirst = reverse_iterator<BidiIt>(last);
const auto rlast = reverse_iterator<BidiIt>(first);auto pivot = next(rfirst);
while(pivot != rlast && *pivot >= *prev(pivot))
{
++pivot;
}//this is the last permute, or the next is the same as the begin one
if(pivot == rlast)
{
reverse(rfirst,rlast);
return false;
}
//find the first num great than pivot
auto change = rfirst;
while(*change<=*pivot)
++change;swap(*change,*pivot);
reverse(rfirst,pivot);
return true;
}
};int main()
{
vector<int> num;
num.push_back(1);
num.push_back(1);
num.push_back(2);Solution myS;
myS.permute(num);
return 0;
}
LeetCode OJ--Permutations II,布布扣,bubuko.com