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
.算法思路:
Dict + LL (1次) + Set (2次)。此题非常类似于LRU, 相当于实习LinkedHashMap。
注意事项:
- 注意两种情况,节点不在Map, 在Map中,对应的LL操作同样是add_to_tail和remove_node不过少了move_to_tail。
- 头尾dummy node,初始化要相连。
- 注意删除顺序,先删map中的entry再删Node。否则会出现NPE。新加入是顺序相反。删除节点要将prev和next赋None
- 区别还有ListNode不需要key,因为输入是单值,不是key-value对。数据结构多了set来记录出现两次及以上的元素,直接忽略
Python代码:
1 | class FirstUnique(TestCases): |
算法分析:
每个操作时间复杂度为O(1)
,空间复杂度O(n)
.