First Unique Number
https://leetcode.com/explore/featured/card/30-day-leetcoding-challenge/531/week-4/3313/
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 toshowFirstUnique
andadd
.
实现一种支持O(1)时间查询最早出现的独特元素的数据结构,能实时插入并且当插入的数据已出现过时,该元素将不再看作独特。O(1)时间查询 => Hash Map,O(1)找最早出现并且能O(1)删除 => Doubly Linked List + Hash Map。可以用Python带的OrderedHashMap实现。为了及时释放已不再是独特的元素node所占的内存,用单独一个hash set记录出现过的数据。
class FirstUnique:
class Node:
def __init__(self, value):
self.pre = self.nxt = None
self.v = value
def __init__(self, nums: List[int]):
self.s, self.d = set(), {}
self.head, self.tail = FirstUnique.Node(-1), FirstUnique.Node(-1)
self.head.nxt = self.tail
self.tail.pre = self.head
for i in nums:
self.add(i)
def showFirstUnique(self) -> int:
return self.head.nxt.v
def add(self, value: int) -> None:
if value not in self.s:
node = FirstUnique.Node(value)
self.d[value] = node
node.pre = self.tail.pre
node.nxt = self.tail
node.pre.nxt = node
self.tail.pre = node
self.s.add(value)
elif value in self.d:
node = self.d[value]
node.nxt.pre = node.pre
node.pre.nxt = node.nxt
del self.d[value]
# Your FirstUnique object will be instantiated and called as such:
# obj = FirstUnique(nums)
# param_1 = obj.showFirstUnique()
# obj.add(value)
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