地址：https://leetcode.com/problems/permutations-ii/

題目：

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

Example:

理解：

和permutation的區別就是需要考慮重複的數字會導致重複的結果。

實現1：

來源：Really easy Java solution, much easier than the solutions with very high vote

class Solution {
public:
	vector<vector<int>> permuteUnique(vector<int>& nums) {
		vector<vector<int>> res;
		vector<int> curr;
		vector<bool> flags(nums.size(), false);
		sort(nums.begin(), nums.end());
		backtracking(nums, res, curr, flags);
		return res;
	}

	void backtracking(vector<int>& nums, vector<vector<int>>& res, vector<int>& curr, vector<bool>& flags) {
		if (nums.size() == curr.size()) {
			res.push_back(curr);
		}
		else {
			for (int i = 0; i < nums.size(); ++i) {
				if (flags[i])
					continue;
				if (i != 0 && nums[i] == nums[i - 1] && !flags[i - 1]) continue;
				curr.push_back(nums[i]);
				flags[i] = true;
				backtracking(nums, res, curr, flags);
				flags[i] = false;
				curr.pop_back();
			}
		}
	}
};
 

實現2：

個人感覺這種更直觀一些，就是把nums存在一個map中，如果某個元素還有就新增並遞迴呼叫，否則不新增。

class Solution {
public:
	vector<vector<int>> permuteUnique(vector<int>& nums) {
		vector<vector<int>> res;
		vector<int> curr;
		unordered_map<int, int> m;
		for (int num : nums) {
			m[num]++;
		}
		backtracking(m, res, curr);
		return res;
	}

	void backtracking(unordered_map<int,int>& m, vector<vector<int>>& res, vector<int>& curr) {
		bool flag = false;
		for (auto &iter : m) {
			if (iter.second != 0) {
				--iter.second;
				curr.push_back(iter.first);
				flag = true;
				backtracking(m, res, curr);
				curr.pop_back();
				++iter.second;
			}
		}
		if (!flag)
			res.push_back(curr);
	}
};
 

實現3：

不是很理解這種實現。。
先存一下吧。本質應該還是為了避免把重複的加入。

class Solution {
public:
	vector<vector<int>> permuteUnique(vector<int>& nums) {
        sort(nums.begin(),nums.end());
		vector<vector<int>> res;
		backtracking(nums, res, 0);
		return res;
	}

	void backtracking(vector<int> nums, vector<vector<int>>& res, int begin) {
		if (begin == nums.size()-1) {
			res.push_back(nums);
		}
		else {
			for (int i = begin; i < nums.size(); ++i) {
				if (i != begin&&nums[i] == nums[begin]) continue;
				swap(nums[begin], nums[i]);
				backtracking(nums, res, begin + 1);
			}
		}
	}
};

實現4：

這種方式比較直觀，使用set避免了重複的加入。

class Solution {
public:
	vector<vector<int>> permuteUnique(vector<int>& nums) {
		sort(nums.begin(), nums.end());
		vector<vector<int>> res;
		backtracking(nums, res, 0);
		return res;
	}

	void backtracking(vector<int>& nums, vector<vector<int>>& res, int begin) {
		if (begin >= nums.size()) {
			res.push_back(nums);
		}
		else {
			unordered_set<int> s;
			for (int i = begin; i < nums.size(); ++i) {
				if (s.count(nums[i])) continue;
				s.insert(nums[i]);
				swap(nums[begin], nums[i]);
				backtracking(nums, res, begin + 1);
				swap(nums[begin],nums[i]);
			}
		}
	}
};