[leetcode]380. Insert Delete GetRandom O(1)常数时间插入删除取随机值

Design a data structure that supports all following operations in average O(1) time.

  1. insert(val): Inserts an item val to the set if not already present.
  2. remove(val): Removes an item val from the set if present.
  3. getRandom: Returns a random element from current set of elements. Each element must have the same probability of being returned.

Example:

// Init an empty set.
RandomizedSet randomSet = new RandomizedSet();

// Inserts 1 to the set. Returns true as 1 was inserted successfully.
randomSet.insert(1);

// Returns false as 2 does not exist in the set.
randomSet.remove(2);

// Inserts 2 to the set, returns true. Set now contains [1,2].
randomSet.insert(2);

// getRandom should return either 1 or 2 randomly.
randomSet.getRandom();

// Removes 1 from the set, returns true. Set now contains [2].
randomSet.remove(1);

// 2 was already in the set, so return false.
randomSet.insert(2);

// Since 2 is the only number in the set, getRandom always return 2.
randomSet.getRandom();

Solution1:  HashMap + ArrayList

1. use an ArrayList together with HashMap, making hashmap save key(item)->value(idx) 

2. To reverse item in O(1), avoiding traversal the whole arrayList in O(n) time, we swap the toDelete item with last item in the list 

3. Go through an example like this: 

   use HashMap to get removeIdx:

   Set the last item in the list as replaceValue, why the last item? In order to maintain other indices.

   Deal with HashMap: (1) put  (2)delete 

 Deal with List: (1) set  (2)delete

                                                                                      

 code:

 1 class RandomizedSet {
 2     Map<Integer, Integer> map;
 3     List<Integer> list;
 4     Random r; // java 自带类
 5 
 6     /** Initialize your data structure here. */
 7     public RandomizedSet() {
 8         list = new ArrayList<>();
 9         map = new HashMap<>();
10         r = new Random();    
11     }
12     
13     /** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
14     public boolean insert(int val) {
15         if (map.containsKey(val)) {
16             return false;
17         }
18         map.put(val, list.size());
19         list.add(val);
20         return true;
21     }
22     
23     /** Removes a value from the set. Returns true if the set contained the specified element. */
24     public boolean remove(int val) {
25         /**
26         hashmap
27         30 - 0
28         40 - 1
29         50 - 2
30         60 - 3 
31         
32         list: 30 40  50 60 
33                0   1  2  3
34         **/
35         if (!map.containsKey(val))     return false;
36         int removeIdx = map.get(val);  // Idx: 1
37         int replaceValue = list.get(list.size()-1); // replaceValue : 60 
38         
39         // deal with map
40         map.put(replaceValue, removeIdx);
41         map.remove(val);
42         
43         // deal with list
44         list.set(removeIdx, replaceValue); 
45         list.remove(list.size() -1);
46       
47         return true;     
48     }
49     
50     /** Get a random element from the set. */
51     public int getRandom() {
52         return list.get(r.nextInt(list.size()));
53     }
54 }
原文地址:https://www.cnblogs.com/liuliu5151/p/9841061.html