leetcode 47. Permutations II (生成全排列，有重复元素)

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

思路：和leet code 46 类似，最后用set去个重即可。。

/* ***********************************************
Author :111qqz
Created Time :2017年04月13日 星期四 15时00分48秒
File Name :47.cpp
************************************************ */
class Solution {

public:
    void solve( vector<int>&nums)
    {
	int n = nums.size();
	if (n==0) return;
	int k = -1;
	for ( int i = n-2 ; i >= 0 ; i--)
	{
	    if (nums[i]<nums[i+1])
	    {
		k = i;
		break;
	    }
	}
	if (k==-1)
	{
	    reverse(nums.begin(),nums.end());
	    return;
	}
	int l = -1;
	for ( int i = n-1 ; i >k ; i--)
	{
	    if (nums[k]<nums[i])
	    {
		l =  i;
		break;
	    }
	}
	swap(nums[l],nums[k]);
	reverse(nums.begin()+k+1,nums.end());
    }
    
    void pr (vector<int> &nums)
    {
	int siz = nums.size();
	for ( int i = 0 ; i < siz;  i++)
	    printf("%d%c",nums[i],i==siz-1?'\n':' ');
    }
    vector<vector<int>> permuteUnique(vector<int>& nums) {
	set<vector<int> >se;
	vector<vector<int> >res;
	int n = nums.size();
	int total = 1 ;
	for ( int i = 2 ; i <= n ; i++) total*=i;
	
	for ( int i = 1 ; i <= total ; i++)
	{
	    se.insert(nums);
//	    pr(nums);
	    solve(nums);
	}
	for ( auto &it :se)
	{
	    res.push_back(it);
	}
	return res;
    }

};