【Leetcode】 381. Insert Delete GetRandom O(1) - Duplicates allowed

Description

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

Note

Duplicate elements are allowed.
insert(val): Inserts an item val to the collection.
remove(val): Removes an item val from the collection if present.
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();

Solution

public class RandomizedCollection {

    List<Integer> nums;
    HashMap<Integer, Set<Integer>> map;
    java.util.Random rand = new java.util.Random();

    public RandomizedCollection() {
        nums = new ArrayList<>();
        map = new HashMap<>();
    }

    public boolean insert(int val) {
        boolean contain = map.containsKey(val);
        if (!contain) {
            map.put(val, new LinkedHashSet<>());
        }
        map.get(val).add(nums.size());
        nums.add(val);
        return !contain;
    }

    public boolean remove(int val) {
        boolean contain = map.containsKey(val);
        if (!contain) {
            return false;
        }
        int index = map.get(val).iterator().next();
        map.get(val).remove(index);
        if (index < nums.size() - 1) {
            int lastone = nums.get(nums.size() - 1);
            map.get(lastone).remove(nums.size() - 1);
            map.get(lastone).add(index);
            nums.set(index, lastone);
        }
        nums.remove(nums.size() - 1);
        if (map.get(val).size() == 0) {
            map.remove(val);
        }
        return true;
    }


    public int getRandom() {
        return nums.get(rand.nextInt(nums.size()));
    }
}

Analyse

这道题很值得细细思考，首先，为什么要用LinkedHashSet，而不是普通的set，因为HashSet不能保证插入的顺序，LinkedHashSet可以按照插入的顺序来对数据进行操作。至于为什么要保持顺序，那就是和nums对应，打一个简单的比方。

//map 0 <1>, 1 <1, 2>, 2 <3, 4>
//nums {0, 1, 1, 2, 2>
//remove 1之后，后续的格式为
//map 0 <1>, 1 <2>, 2 <3, 1>
//nums {0, 2, 1, 2}
//这样数字对应的index和map里的数字的LinkedHashSet是一样的。
//为什么需要对应一样，那接下来看
//要是remove2很直观，如果接下来remove1
//map 0 <1>, 1 <>, 2 <3, 1>
//这样接下来操作
//map 0 <1>, 1 <>, 2 <2, 1>
//一套技能下来，对数据结构的理解也更深