资讯专栏INFORMATION COLUMN

[LeetCode] 432. All O`one Data Structure

tanglijun / 2710人阅读

Problem

Implement a data structure supporting the following operations:

Inc(Key) - Inserts a new key with value 1. Or increments an existing key by 1. Key is guaranteed to be a non-empty string.
Dec(Key) - If Key"s value is 1, remove it from the data structure. Otherwise decrements an existing key by 1. If the key does not exist, this function does nothing. Key is guaranteed to be a non-empty string.
GetMaxKey() - Returns one of the keys with maximal value. If no element exists, return an empty string "".
GetMinKey() - Returns one of the keys with minimal value. If no element exists, return an empty string "".
Challenge: Perform all these in O(1) time complexity.

Note
Use DoublyLinkedList to get max/min key (max/min count)
We put max count Bucket at tail, min count Bucket at head, O(1)

So the node(Bucket) must contain count property
And we can use map  to access each node in O(1)
Each Bucket could have multiple keys of same count, so use Set to store keys

To get count of a key, use Map  to achieve O(1)

Solution
class AllOne {
    private class Bucket {
        int count;
        Set keySet;
        Bucket prev;
        Bucket next;
        public Bucket(int count) {
            this.count = count;
            this.keySet = new HashSet<>();
        }
    }

    private Bucket head;
    private Bucket tail;
    private Map countBucketMap;
    private Map keyCountMap;
    
    public AllOne() {
        head = new Bucket(Integer.MIN_VALUE);
        tail = new Bucket(Integer.MAX_VALUE);
        head.next = tail;
        tail.prev = head;
        countBucketMap = new HashMap<>();
        keyCountMap = new HashMap<>();
    }
    
    public void inc(String key) {
        if (keyCountMap.containsKey(key)) {
            //existing key
            updateKey(key, 1);
        } else {
            //new key
            keyCountMap.put(key, 1);
            if (head.next.count != 1) {
                addBucketAfter(head, new Bucket(1));
            }
            head.next.keySet.add(key);
            countBucketMap.put(1, head.next);
        }
    }
    
    public void dec(String key) {
        if (keyCountMap.containsKey(key)) {
            int count = keyCountMap.get(key);
            if (count == 1) {
                keyCountMap.remove(key);
                removeKeyFromBucket(countBucketMap.get(count), key);
            } else {
                updateKey(key, -1);
            }
        }
    }
    
    private void updateKey(String key, int i) {
        int count = keyCountMap.get(key);
        keyCountMap.put(key, count+i);
        Bucket pre = countBucketMap.get(count);
        Bucket cur;
        if (countBucketMap.containsKey(count+i)) {
            cur = countBucketMap.get(count+i);
        } else {
            cur = new Bucket(count+i);
            countBucketMap.put(count+i, cur);
            if (i == 1) {
                addBucketAfter(pre, cur);
            } else {
                addBucketAfter(pre.prev, cur);
            }
        }
        cur.keySet.add(key);
        removeKeyFromBucket(pre, key);
    }
    
    private void addBucketAfter(Bucket pre, Bucket cur) {
        Bucket next = pre.next;
        pre.next = cur;
        cur.prev = pre;
        cur.next = next;
        next.prev = cur;
    }
    
    private void removeKeyFromBucket(Bucket bucket, String key) {
        bucket.keySet.remove(key);
        if (bucket.keySet.size() == 0) {
            bucket.prev.next = bucket.next;
            bucket.next.prev = bucket.prev;
            bucket.prev = null;
            bucket.next = null;
            countBucketMap.remove(bucket.count);
        }
    }
    
    public String getMaxKey() {
        return tail.prev == head ? "" : (String) tail.prev.keySet.iterator().next();
    }
    
    public String getMinKey() {
        return head.next == tail ? "" : (String) head.next.keySet.iterator().next();
    }
}

Reference: https://leetcode.com/problems...

文章版权归作者所有,未经允许请勿转载,若此文章存在违规行为,您可以联系管理员删除。

转载请注明本文地址:https://www.ucloud.cn/yun/72309.html

相关文章

  • [LeetCode]Find Median from Data Stream

    Find Median from Data Stream Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value. Examp...

    suemi 评论0 收藏0
  • leetcode-211-Add and Search Word - Data structure

    摘要:原题总结栈的利用,先进后出的作用,可以保持链表一类的数据的连贯操作,可以用来替代广度搜索。每一层次可以用进栈出栈进行替代。形式的数据结构,有记忆状态的作用。应用字符串的遍历,广度搜索。 原题: 211. Add and Search Word - Data structure design Design a data structure that supports the follo...

    Alliot 评论0 收藏0
  • [LeetCode] 348. Design Tic-Tac-Toe

    Problem Design a Tic-tac-toe game that is played between two players on a n x n grid. You may assume the following rules: A move is guaranteed to be valid and is placed on an empty block.Once a winnin...

    MobService 评论0 收藏0
  • [Leetcode] Find Median from Data Stream 数据流中位数

    摘要:最大堆存的是到目前为止较小的那一半数,最小堆存的是到目前为止较大的那一半数,这样中位数只有可能是堆顶或者堆顶两个数的均值。我们将新数加入堆后,要保证两个堆的大小之差不超过。最大堆堆顶大于新数时,说明新数将处在所有数的下半部分。 Data Stream Median 最新更新:https://yanjia.me/zh/2019/02/... Median is the middle v...

    heartFollower 评论0 收藏0
  • [LeetCode] 211. Add and Search Word - Data structu

    Problem Design a data structure that supports the following two operations: void addWord(word)bool search(word)search(word) can search a literal word or a regular expression string containing only let...

    mozillazg 评论0 收藏0

发表评论

0条评论

最新活动
阅读需要支付1元查看
<