Day 28 of 30 days Leetcode Challenge - 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 toshowFirstUnique
and add.
Solution
It's similar to LRU cache
Use a hash table and a Doubly Linked List
class FirstUnique {
private var count = [Int: Int]()
private var nodes = [Int: Node]()
private var firstNode: Node?
private var lastNode: Node?
init(_ nums: [Int]) {
for val in nums {
add(val)
}
}
func showFirstUnique() -> Int {
if let node = firstNode {
return node.val
}
return -1
}
func add(_ value: Int) {
count[value] = (count[value] ?? 0) + 1
if count[value] == 1 {
var node = Node(value)
lastNode?.next = node
node.prev = lastNode
lastNode = node
if nodes.count == 0 {
firstNode = node
}
nodes[value] = node
}else{
var node = nodes[value]
if firstNode == node {
firstNode = node?.next
}
if lastNode == node {
lastNode = node?.prev
}
node?.prev?.next = node?.next
node?.next?.prev = node?.prev
nodes[value] = nil
}
}
}
class Node : Equatable {
let val: Int
var next: Node?
var prev: Node?
init(_ val: Int) {
self.val = val
}
static func ==(lhs: Node, rhs: Node) -> Bool {
return lhs.val == rhs.val
}
}