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

Note: Duplicate elements are allowed.

1. insert(val): Inserts an item val to the collection.
2. remove(val): Removes an item val from the collection if present.
3. getRandom: Returns a random element from current collection of elements. The probability of each element being returned is linearly related to the number of same value the collection contains.

Example:

// Init an empty collection.
RandomizedCollection collection = new RandomizedCollection();

// Inserts 1 to the collection. Returns true as the collection did not contain 1.
collection.insert(1);

// Inserts another 1 to the collection. Returns false as the collection contained 1. Collection now contains [1,1].
collection.insert(1);

// Inserts 2 to the collection, returns true. Collection now contains [1,1,2].
collection.insert(2);

// getRandom should return 1 with the probability 2/3, and returns 2 with the probability 1/3.
collection.getRandom();

// Removes 1 from the collection, returns true. Collection now contains [1,2].
collection.remove(1);

// getRandom should return 1 and 2 both equally likely.
collection.getRandom();

題目

思路

程式碼

 1 class RandomizedCollection {
 2                      // TreeSet用來處理每個相同val的不同idx
 3     HashMap<Integer, TreeSet<Integer>> map;
 4     List<Integer> list;
 5     Random random;
 6     /** Initialize your data structure here. */
 7     public RandomizedCollection() {
 
 8         map = new HashMap<>();
 9         list = new ArrayList<>();
10         random = new Random();
11     }
12 
13     /** Inserts a value to the collection. Returns true if the collection did not already contain the specified element. */
14     public boolean insert(int val) {
15         TreeSet<Integer> idxSet = map.get(val);
16         // 如果該val不在map裡，則先扔到map裡一對新pair
17         if (idxSet == null) {
18             idxSet = new TreeSet<>();
19             map.put(val, idxSet);
20         }
21         //如果該val在map裡，則給該val分配一個新idx，記錄在idxSet中
22         // list.size() 這個操作的是O（1）的，原始碼可查
23         idxSet.add(list.size());
24         list.add(val);
25         //if the collection did not already contain，則新加一個idx後 idxSet.size() 為1
26         // if the collection already contain，則新加一個idx後 idxSet.size() >1
27         return idxSet.size() == 1;
28     }
29 
30     /** Removes a value from the collection. Returns true if the collection contained the specified element. */
31     public boolean remove(int val) {
32         TreeSet<Integer> idxSet = map.get(val);  // val(20) : idxSet(0-3-6)
33         // corner case
34         if (idxSet == null || idxSet.isEmpty()) {
35             return false;
36         }
37         // removeIdx可以在idxSet找任意idex，pollLast()只是方便起見
38         int removeIdx = idxSet.pollLast(); 
39         int replaceVal = list.get(list.size() - 1); //Tail of list
40         // 因為允許duplicates, 這個TreeSet裡裝著所有replaceVal對應的不同idx
41         TreeSet<Integer> tailIdxSet = map.get(replaceVal);
42         if (replaceVal != val) {
43             tailIdxSet.pollLast(); //Remove last removeIdx of list replaceVal
44             tailIdxSet.add(removeIdx); //Add removeIdx to replaceVal removeIdx set
45             list.set(removeIdx, replaceVal);
46         }
47         list.remove(list.size() - 1);
48         return true;
49     }
50 
51     /** Get a random element from the collection. */
52     public int getRandom() {
53         return list.get(random.nextInt(list.size()));
54     }
55 }