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:
- 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.
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.