zoukankan      html  css  js  c++  java
  • LeetCode 432. All O`one Data Structure

    原题链接在这里:https://leetcode.com/problems/all-oone-data-structure/?tab=Description

    题目:

    Implement a data structure supporting the following operations:

    1. 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.
    2. 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.
    3. GetMaxKey() - Returns one of the keys with maximal value. If no element exists, return an empty string "".
    4. 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.

    题解:

    很像LFU的机制.

    用Bucket组成一个double linked list. 按照Bucket的count从小到大排序.

    inc 把key move到下一个Bucket.

    dec把key move到前一个Bucket.

    min key就在前面, head.next.

    max key就出现在后面, tail.pre.

    Time Complexity: inc, dec, getaMaxKey, getMinKey, O(1).

    Space: O(n). n是现有key的个数.

    AC Java:

      1 public class AllOne {
      2     Bucket head;
      3     Bucket tail;
      4     Map<String, Integer> keyCount;
      5     Map<Integer, Bucket> countBucket;
      6     
      7     /** Initialize your data structure here. */
      8     public AllOne() {
      9         head = new Bucket(-1);
     10         tail = new Bucket(-1);
     11         head.next = tail;
     12         tail.pre = head;
     13         keyCount = new HashMap<String, Integer>();
     14         countBucket = new HashMap<Integer, Bucket>();
     15     }
     16     
     17     /** Inserts a new key <Key> with value 1. Or increments an existing key by 1. */
     18     public void inc(String key) {
     19         if(keyCount.containsKey(key)){
     20             moveKey(key, 1);
     21         }else{
     22             keyCount.put(key, 1);
     23             if(head.next.count != 1){
     24                 addBucketAfter(new Bucket(1), head);
     25             }
     26             head.next.keySet.add(key);
     27             countBucket.put(1, head.next);
     28         }
     29     }
     30     
     31     /** Decrements an existing key by 1. If Key's value is 1, remove it from the data structure. */
     32     public void dec(String key) {
     33         if(keyCount.containsKey(key)){
     34             int count = keyCount.get(key);
     35             if(count == 1){
     36                 keyCount.remove(key);
     37                 removeKeyFromBucket(countBucket.get(count), key);
     38             }else{
     39                 moveKey(key, -1);
     40             }
     41         }
     42     }
     43     
     44     /** Returns one of the keys with maximal value. */
     45     public String getMaxKey() {
     46         return tail.pre == head ? "" : tail.pre.keySet.iterator().next();
     47     }
     48     
     49     /** Returns one of the keys with Minimal value. */
     50     public String getMinKey() {
     51         return head.next == tail ? "" : head.next.keySet.iterator().next();
     52     }
     53     
     54     private void moveKey(String key, int offset){
     55         int count = keyCount.get(key);
     56         keyCount.put(key, count+offset);
     57         Bucket cur = countBucket.get(count);
     58         Bucket moveToBucket;
     59         if(countBucket.containsKey(count+offset)){
     60             moveToBucket = countBucket.get(count+offset);
     61         }else{
     62             moveToBucket = new Bucket(count+offset);
     63             countBucket.put(count+offset, moveToBucket);
     64             addBucketAfter(moveToBucket, offset == 1 ? cur : cur.pre);
     65         }
     66         
     67         moveToBucket.keySet.add(key);
     68         removeKeyFromBucket(cur, key);
     69     }
     70     
     71     private void addBucketAfter(Bucket newBucket, Bucket preBucket){
     72         newBucket.next = preBucket.next;
     73         newBucket.pre = preBucket;
     74         preBucket.next.pre = newBucket;
     75         preBucket.next = newBucket;
     76     }
     77     
     78     private void removeKeyFromBucket(Bucket cur, String key){
     79         cur.keySet.remove(key);
     80         if(cur.keySet.size() == 0){
     81             countBucket.remove(cur.count);
     82             cur.pre.next = cur.next;
     83             cur.next.pre = cur.pre;
     84         }
     85     }
     86 }
     87 
     88 class Bucket{
     89     int count;
     90     Set<String> keySet;
     91     Bucket pre;
     92     Bucket next;
     93     
     94     public Bucket(int count){
     95         this.count = count;
     96         keySet = new LinkedHashSet<String>();
     97     }
     98 }
     99 
    100 /**
    101  * Your AllOne object will be instantiated and called as such:
    102  * AllOne obj = new AllOne();
    103  * obj.inc(key);
    104  * obj.dec(key);
    105  * String param_3 = obj.getMaxKey();
    106  * String param_4 = obj.getMinKey();
    107  */

    类似LFU Cache.

  • 相关阅读:
    weui-switch开关控件,表单提交后如何取值
    [转]判断存储过程、触发器、视图是否存在并删除
    修改服务器的3389端口
    如何在react-native 中优雅的使用 redux
    react native js 与 native 的通信与交互方式
    对 JS virtual DOM 的一知半解
    Python的实例方法,类方法,静态方法之间的区别及调用关系
    redux 管理你的 react 应用
    react- native 入门
    git 指南
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/6491294.html
Copyright © 2011-2022 走看看