LeetCode 1429 First Unique Number

LeetCode

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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:

<pre>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 </pre>

Example 2:

<pre>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 </pre>

Example 3:

<pre>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 </pre>

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.

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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。

由易到难

题目	LL	Map	其他数据结构
LRU	按时间顺序，头删尾入	值->LL节点	N/A
First Unique	按时间顺序，任意删尾入	值->LL节点	Set存两次以上
LFU	多个时间顺序的LL	值->LL节点	第二个Map存freq-> LL的首节点以及min_freq

注意事项：

1. 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。
2. 头尾dummy node，初始化要相连。
3. 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
4. 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：

class FirstUnique(TestCases):

    '''
    2, 3, 5, 5
    dict:
    {2:1
     3:1
     5:2 (remove)
     }
    LinkedList: 2, 3, 5(r), 5(r)
    dict:
    {
        2: Node(2)
        3: Node(3)
        5: Node(5)
    }
    
    '''

    def __init__(self, nums: List[int]):
        self.head = ListNode(0)  # only store unique elements
        self.tail = ListNode(0)
        self.key_to_node = {}  # only store unique elements
        self.non_unique_set = set()
        self.head.next, self.tail.prev = self.tail, self.head

        for n in nums:
            self.add(n)

    def showFirstUnique(self) -> int:
        if len(self.key_to_node) == 0:
            return -1
        return self.head.next.val
    # 2, 3, 5, 5
    def add(self, value: int) -> None:
        if value in self.non_unique_set:
            return

        if value not in self.key_to_node:
            new_node = ListNode(value)
            self.add_to_tail(new_node)
            self.key_to_node[value] = new_node
        else:
            to_be_removed_node = self.key_to_node[value]
            self.key_to_node.pop(value)
            self.remove_node(to_be_removed_node)
            self.non_unique_set.add(value)

    def add_to_tail(self, node):
        predecessor = self.tail.prev
        predecessor.next, node.prev = node, predecessor
        node.next, self.tail.prev = self.tail, node

    def remove_node(self, node):
        predecessor, successor = node.prev, node.next
        predecessor.next, successor.prev = successor, predecessor
        node.prev, node.next = None, None


class ListNode:

    def __init__(self, val, next = None, prev = None):
        self.val = val
        self.next = next
        self.prev = prev

算法分析：

每个操作时间复杂度为O(1)，空间复杂度O(n).