[Leetcode] Permutations II Arrangement

Given A collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2]The following unique permutations:
[1,1,2], [1,2,1] , and [2,1,1] .

Show TagsBacktracking This is the input an array, there may be duplicates, the output of all the arrangement. The question called STL is as follows:

1#include <vector>2#include <iterator>3#include <algorithm>4#include <iostream>5 using namespacestd;6 7 classSolution {8  Public:9vector<vector<int> > Permuteunique (vector<int> &num) {Ten         intn =num.size (); Onevector<vector<int> >ret; A         if(n<1)returnret; -         if(n<2) {ret.push_back (num);returnret;} - sort (Num.begin (), Num.end ()); the ret.push_back (num); -          while(Next_permutation (Num.begin (), Num.end ())) { - ret.push_back (num); -         } +         returnret; -     } + }; A  at intMain () - { -vector<int> num = {1,1,2}; - solution Sol; -vector<vector<int> > RET =sol.permuteunique (num); -      for(intI=0; I<ret.size (); i++){ inCopy (Ret[i].begin (), Ret[i].end (),ostream_iterator<int> (cout," ")); -cout<<Endl; to     } +     return 0; -}

View Code

If you do not call, is to write a next_permutation, in permutations wrote a good many versions, review the implementation of the STL logic it:

1. Input array a[], from right-to-left traversal, looking for adjacent two numbers, so that left<mid, not found? Just no, return false.
2. Again from the right to the left to traverse, to look for >left, this time you do not need to be connected to the back, because there are 1 of this, so there is a guaranteed value (mid)
3. Swap left and right.
4. Reverse mid and its right.
5. End.

[Leetcode] Permutations II Arrangement