Leetcode: All O`one Data Structure
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.
Solution: O(1) time complexity
解题思路主要参考了网友ivancjw的帖子，数据结构参考了https://discuss.leetcode.com/topic/65634/java-ac-all-strict-o-1-not-average-o-1-easy-to-read用bucket，思路是，我们建立一个次数分层的结构，次数多的在顶层，每一层放相同次数的key值，例如下面这个例子：
"A": 4, "B": 4, "C": 2, "D": 1
那么用我们设计的结构保存出来就是：
row0: val = 4, keys = {"A", "B"}
row1: val = 2, keys = {"C"}
row2: val = 1, keys = {"D"}
1 public class AllOne { 2 public class Bucket { 3 int count; 4 Bucket prev; 5 Bucket next; 6 HashSet<String> keySet; 7 public Bucket(int num) { 8 this.count = num; 9 this.keySet = new HashSet<String>(); 10 } 11 } 12 13 Bucket head; 14 Bucket tail; 15 HashMap<String, Integer> keyCountMap; 16 HashMap<Integer, Bucket> countBucketMap; 17 18 19 /** Initialize your data structure here. */ 20 public AllOne() { 21 this.head = new Bucket(Integer.MIN_VALUE); 22 this.tail = new Bucket(Integer.MAX_VALUE); 23 head.next = tail; 24 tail.prev = head; 25 this.keyCountMap = new HashMap<String, Integer>(); 26 this.countBucketMap = new HashMap<Integer, Bucket>(); 27 } 28 29 /** Inserts a new key <Key> with value 1. Or increments an existing key by 1. */ 30 public void inc(String key) { 31 if (keyCountMap.containsKey(key)) { 32 change(key, 1); 33 } 34 else { 35 keyCountMap.put(key, 1); 36 if (head.next.count != 1) { //dont have the 1 bucket 37 addBucketAfter(new Bucket(1), head); 38 countBucketMap.put(1, head.next); 39 } 40 head.next.keySet.add(key); 41 } 42 } 43 44 /** Decrements an existing key by 1. If Key's value is 1, remove it from the data structure. */ 45 public void dec(String key) { 46 if (keyCountMap.containsKey(key)) { 47 int count = keyCountMap.get(key); 48 if (count == 1) { 49 keyCountMap.remove(key); 50 removeKeyFromBucket(countBucketMap.get(count), key); 51 } 52 else change(key, -1); 53 } 54 } 55 56 /** Returns one of the keys with maximal value. */ 57 public String getMaxKey() { 58 return tail.prev==head? "" : (String)tail.prev.keySet.iterator().next(); 59 } 60 61 /** Returns one of the keys with Minimal value. */ 62 public String getMinKey() { 63 return head.next==tail? "" : (String)head.next.keySet.iterator().next(); 64 } 65 66 public void change(String key, int offset) { 67 //get count, update keyCountMap 68 int count = keyCountMap.get(key); 69 keyCountMap.put(key, count+offset); 70 71 //get current bucket 72 Bucket curBucket = countBucketMap.get(count); 73 74 //new bucket 75 Bucket newBucket; 76 if (countBucketMap.containsKey(count+offset)) { 77 newBucket = countBucketMap.get(count+offset); 78 } 79 else { 80 newBucket = new Bucket(count+offset); 81 countBucketMap.put(count+offset, newBucket); 82 addBucketAfter(newBucket, (offset==1? curBucket : curBucket.prev)); 83 } 84 newBucket.keySet.add(key); 85 removeKeyFromBucket(curBucket, key); 86 } 87 88 public void removeKeyFromBucket(Bucket cur, String key) { 89 cur.keySet.remove(key); 90 if (cur.keySet.size() == 0) { 91 removeBucketFromList(cur); 92 countBucketMap.remove(cur.count); 93 } 94 } 95 96 public void removeBucketFromList(Bucket cur) { 97 cur.prev.next = cur.next; 98 cur.next.prev = cur.prev; 99 cur.next = null; 100 cur.prev = null; 101 } 102 103 public void addBucketAfter(Bucket bucket, Bucket preBucket) { 104 bucket.prev = preBucket; 105 bucket.next = preBucket.next; 106 preBucket.next.prev = bucket; 107 preBucket.next = bucket; 108 } 109 } 110 111 /** 112 * Your AllOne object will be instantiated and called as such: 113 * AllOne obj = new AllOne(); 114 * obj.inc(key); 115 * obj.dec(key); 116 * String param_3 = obj.getMaxKey(); 117 * String param_4 = obj.getMinKey(); 118 */
Solution 2: 如果不要求O（1）time, 这个用两个heap方法很常规
1 public class AllOne { 2 3 class Node{ 4 String key; 5 int val; 6 public Node(String key, int val) { 7 this.key = key; 8 this.val = val; 9 } 10 } 11 /** Initialize your data structure here. */ 12 HashMap<String, Node> map; 13 PriorityQueue<Node> minQ; 14 PriorityQueue<Node> maxQ; 15 public AllOne() { 16 map = new HashMap<String, Node>(); 17 minQ = new PriorityQueue<Node>(new Comparator<Node>(){ 18 public int compare(Node a, Node b) { 19 return a.val - b.val; 20 } 21 }); 22 maxQ = new PriorityQueue<Node>(new Comparator<Node>(){ 23 public int compare(Node a, Node b) { 24 return b.val - a.val; 25 } 26 }); 27 } 28 29 /** Inserts a new key <Key> with value 1. Or increments an existing key by 1. */ 30 public void inc(String key) { 31 if (!map.containsKey(key)) { 32 map.put(key, new Node(key, 1)); 33 Node node = map.get(key); 34 minQ.add(node); 35 maxQ.add(node); 36 } else { 37 Node node = map.get(key); 38 minQ.remove(node); 39 maxQ.remove(node); 40 node.val++; 41 map.put(key, node); 42 minQ.add(node); 43 maxQ.add(node); 44 } 45 } 46 47 /** Decrements an existing key by 1. If Key's value is 1, remove it from the data structure. */ 48 public void dec(String key) { 49 if (map.containsKey(key)) { 50 Node node = map.get(key); 51 if (node.val == 1) { 52 map.remove(key); 53 minQ.remove(node); 54 maxQ.remove(node); 55 } else { 56 minQ.remove(node); 57 maxQ.remove(node); 58 node.val--; 59 map.put(key, node); 60 minQ.add(node); 61 maxQ.add(node); 62 } 63 } 64 } 65 66 /** Returns one of the keys with maximal value. */ 67 public String getMaxKey() { 68 return maxQ.isEmpty() ? "" : maxQ.peek().key; 69 } 70 71 /** Returns one of the keys with Minimal value. */ 72 public String getMinKey() { 73 return minQ.isEmpty() ? "" : minQ.peek().key; 74 } 75 }
