# [47. Permutations II](https://leetcode.com/problems/permutations-ii/)
# 思路
有了[31. Next Permutation](https://leetcode.com/problems/next-permutation/)和[46. Permutations](https://leetcode.com/problems/permutations/)的做题经验，这题就显得是一个送分题了。     
这题与46题唯一区别就是此题允许重复元素，所以需要注意在元素比较的时候相等的情况。参考[46的题解](https://github.com/ShusenTang/LeetCode/blob/master/solutions/46.%20Permutations.md)即可很容易写出代码。


# C++
## 使用STL中的next_permutations(完全同46题)
``` C++
class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>>res;
        if(nums.empty()) return res;
        sort(nums.begin(), nums.end()); // 先排序
        res.push_back(nums);
        while(next_permutation(nums.begin(), nums.end())) res.push_back(nums);
        return res;
    }
};
```
## 手动实现
``` C++
class Solution {
private:
    bool my_next_permute(vector<int>& nums){
        int len = nums.size();
        int i = len - 1;
        while(i > 0 && nums[i] <= nums[i - 1]) i--; // 与46题的不同点一
        if(i == 0) return false;
        
        int mid, low = i, high = len - 1;
        while(low <= high){ // 二分查找
            mid = low + (high - low) / 2;
            if(nums[mid] <= nums[i - 1]) high = mid - 1;  // 与46题的不同点二
            else low = mid + 1;
        }
        
        // int high = len - 1;
        // while(nums[i - 1] >= nums[high]) high--; // 与46题的不同点二
        
        swap(nums[i - 1], nums[high]);
        reverse(nums.begin() + i, nums.end());
        return true;
    }
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>>res;
        if(nums.empty()) return res;
        sort(nums.begin(), nums.end()); // 先排序
        res.push_back(nums);
        while(my_next_permute(nums)) res.push_back(nums);
        return res;
    }
};
```