LeetCode First Unique Number

You have a queue of integers, you need to retrieve the first unique integer in the queue.

Implement the FirstUnique class:

• FirstUnique(int[] nums) Initializes the object with the numbers in the queue.
• int showFirstUnique() returns the value of the first unique integer of the queue, and returns -1 if there is no such integer.
• void add(int value) insert value to the queue.

Example 1:

Input: 
["FirstUnique","showFirstUnique","add","showFirstUnique","add","showFirstUnique","add","showFirstUnique"]
[[[2,3,5]],[],[5],[],[2],[],[3],[]]
Output: 

[null,2,null,2,null,3,null,-1]

Explanation: FirstUnique firstUnique = new FirstUnique([2,3,5]); firstUnique.showFirstUnique(); // return 2 firstUnique.add(5); // the queue is now [2,3,5,5] firstUnique.showFirstUnique(); // return 2 firstUnique.add(2);            // the queue is now [2,3,5,5,2] firstUnique.showFirstUnique(); // return 3 firstUnique.add(3);            // the queue is now [2,3,5,5,2,3] firstUnique.showFirstUnique(); // return -1

Example 2:

Input: 
["FirstUnique","showFirstUnique","add","add","add","add","add","showFirstUnique"]
[[[7,7,7,7,7,7]],[],[7],[3],[3],[7],[17],[]]
Output: 

[null,-1,null,null,null,null,null,17]

Explanation: FirstUnique firstUnique = new FirstUnique([7,7,7,7,7,7]); firstUnique.showFirstUnique(); // return -1 firstUnique.add(7); // the queue is now [7,7,7,7,7,7,7] firstUnique.add(3);            // the queue is now [7,7,7,7,7,7,7,3] firstUnique.add(3);            // the queue is now [7,7,7,7,7,7,7,3,3] firstUnique.add(7);            // the queue is now [7,7,7,7,7,7,7,3,3,7] firstUnique.add(17);           // the queue is now [7,7,7,7,7,7,7,3,3,7,17] firstUnique.showFirstUnique(); // return 17

Example 3:

Input: 
["FirstUnique","showFirstUnique","add","showFirstUnique"]
[[[809]],[],[809],[]]
Output: 

[null,809,null,-1]

Explanation: FirstUnique firstUnique = new FirstUnique([809]); firstUnique.showFirstUnique(); // return 809 firstUnique.add(809); // the queue is now [809,809] firstUnique.showFirstUnique(); // return -1

Constraints:

• 1 <= nums.length <= 10^5
• 1 <= nums[i] <= 10^8
• 1 <= value <= 10^8
• At most 50000 calls will be made to showFirstUnique and add.

---

设计题。设计一个队列，能够快速返回当前队列中第一个unique的数。

和之前的LRU很类似，使用list+unordered_map实现。list保存当前unique的数，unordered_map保存每个unique的数在list中的迭代器。当插入一个数到队列中时，首先看看在不在unordered_map中，如果不在，说明这个数是unique的，插入list和unordered_map。如果在，则看看unordered_map中的迭代器，如果迭代器不为空，说明这个数之前是unique的，需要从list中删掉；否则说明这个数之前就已经不是unique的了，不做任何操作。

完整代码如下：

class FirstUnique {
private:
	list<int> uniques_;
	unordered_map<int, list<int>::iterator> hash;
public:
	FirstUnique(vector<int>& nums) {
		for (int i = 0; i < nums.size(); ++i) {
			add(nums[i]);
		}
	}

	int showFirstUnique() {
		if (uniques_.empty())return -1;
		else return uniques_.front();
	}

	void add(int value) {
		if (hash.find(value) == hash.end()) {
			uniques_.push_back(value);
			hash[value] = --uniques_.end();
		}
		else {
			if (hash[value] != uniques_.end()) {
				uniques_.erase(hash[value]);
				hash[value] = uniques_.end();
			}
		}
	}
};

本代码提交AC，用时520MS。