Insert Delete GetRandom O(1) & Duplicates allowed

380. Insert Delete GetRandom O(1)

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.

思路

HashSet能够实现O(1)时间复杂度的insert和remove，可是要求getRandom也是O(1)，只用HashSet是不能够的。在List里面随机找一个数的时间是O(1)，因此能够用list来记录加进去的value，这样insert和getRandom都是O(1)了。remove的操做比较有意思，在一个list里面找到相应的值以后和最后一个位置上的值调换，这个remove就是常数时间了。可是在list里面查找的时间复杂度依然是O(N)，能够想到用hashmap来记录对应的index，这样查找的时间也是常数。

复杂度

Time: O(1), Space: O(N)

代码

Map<Integer, Integer> map;
    List<Integer> list;
    Random random;
    public RandomizedSet() {
        // hashmap to realize insert, remove in O(1), the value is the index in list
        map = new HashMap();
        // list to realize get random in O(1);
        list = new ArrayList();
        random = new Random();
    }
    
    /** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
    public boolean insert(int val) {
        if(map.containsKey(val)) return false;
        map.put(val, list.size());
        list.add(val);
        return true;
    }
    
    /** Removes a value from the set. Returns true if the set contained the specified element. */
    public boolean remove(int val) {
        // change the val with the last one, then delete last one
        if(!map.containsKey(val)) return false;
        int delete_index = map.get(val);
        int last = list.get(list.size() - 1);
        // change the last element to the delete place
        list.set(delete_index, last);
        // update the index in map
        map.replace(last, delete_index);
        // remove in the list
        list.remove((int) list.size() - 1);
        // remove in the map
        map.remove(val);
        return true;
    }
    
    /** Get a random element from the set. */
    public int getRandom() {
        if(list.size() == 0) return -1;
        return list.get(random.nextInt(list.size()));
    }

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

思路

容许duplication以后麻烦了许多。如今hashmap里面要存全部的index。在remove的时候复杂度要保持O(1)，每次仍是要取最后一个值和删除的值交换，因此须要map里面找到最后一个值，以后删除最大的index的时间复杂度是O(1)，同时把index更新成当前删除的值的index以后，全部的index还要保持顺序。

heap

用heap能够保持顺序，可是poll的时间复杂度是O(logN)。

代码

Map<Integer, PriorityQueue<Integer>> map;
    List<Integer> list;
    Random random;
    public RandomizedCollection() {
        map = new HashMap();
        list = new ArrayList();
        random = new Random();
    }
    
    /** Inserts a value to the collection. Returns true if the collection did not already contain the specified element. */
    public boolean insert(int val) {
        if(map.containsKey(val)) {
            map.get(val).add(list.size());
            list.add(val);
            return false;
        }
        else {
            map.put(val, new PriorityQueue<>((a, b) -> b - a));
            map.get(val).add(list.size());
            list.add(val);
            return true;
        }
    }
    
    /** Removes a value from the collection. Returns true if the collection contained the specified element. */
    public boolean remove(int val) {
        if(!map.containsKey(val)) return false;
        int delete_index = map.get(val).peek();
        int last = list.get(list.size() - 1);
        // update
        list.set(delete_index, last);
        map.get(last).poll();
        map.get(last).add(delete_index);
        // delete
        list.remove(list.size() - 1);
        map.get(val).poll();
        if(map.get(val).size() == 0) map.remove(val);
        return true;
    }
    
    /** Get a random element from the collection. */
    public int getRandom() {
        return list.get(random.nextInt(list.size()));
    }

LinkedHashSet