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题目链接：https://leetcode.com/problems/insert-delete-getrandom-o1-duplicates-allowed/
题目：
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();
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思路：
跟上一题不同点在于允许插入重复元素，我们仍然用list存储插入元素，用hashmap维护元素与下标之间的对应关系，不同点在于我们用一个集合存储某元素所有下标（因为可能有重复元素插入，但它们是按序插入list的即下标是不同的），那么接下来考虑用什么集合存储某元素的所有下标呢？
首先考虑list，但当我们remove某元素时，需要更新某元素的下标集合，此时用list，时间复杂度为O(n)。
考虑set，因为是按序插入list的，即下标是不同的，hashset可以存储，且remove操作时间复杂度为O(1)。
算法：
/** Initialize your data structure here. */
public RandomizedCollection() {
}
List<Integer> vals = new ArrayList<Integer>();
Map<Integer, Set<Integer>> val2idx = new HashMap<Integer, Set<Integer>>();
/** Inserts a value to the collection. Returns true if the collection did not already contain the specified element. */
public boolean insert(int val) {
boolean flag = false;
Set<Integer> idxs = null;
if (val2idx.containsKey(val)) {
idxs = val2idx.get(val);
flag=false;
} else {
idxs = new HashSet<Integer>();
flag=true;
}
idxs.add(vals.size());
vals.add(val);
val2idx.put(val,idxs);
return flag;
}
/** Removes a value from the collection. Returns true if the collection contained the specified element. */
public boolean remove(int val) {
if (!val2idx.containsKey(val)) {
return false;
} else {
Set<Integer> rmIdxs = val2idx.get(val);
int rmIdx = rmIdxs.iterator().next();//被删除元素的下标
if (rmIdx < vals.size() - 1&&val!=vals.get(vals.size()-1)) {//若刪除的不是末尾元素，且被刪除元素不等于末尾元素（若相等的话就当做删除末尾
// 将末尾元素存入被删除元素的位置，并更新相应下标
int lastElem = vals.get(vals.size() - 1);//末尾元素
Set<Integer> lastElemIdxs = val2idx.get(lastElem);//末尾元素的index集合
lastElemIdxs.remove(vals.size()-1);
lastElemIdxs.add(rmIdx);
vals.set(rmIdx,lastElem);
val2idx.put(lastElem,lastElemIdxs);
}
rmIdxs.remove(rmIdx);
if(rmIdxs.size()==0){
val2idx.remove(val);
}else{
val2idx.put(val,rmIdxs);
}
vals.remove(vals.size() - 1);
return true;
}
}
Random r = new Random();
public int getRandom() {
return vals.get(r.nextInt(vals.size()));
}【Leetcode】Insert Delete GetRandom O(1) - Duplicates allowed
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原文地址：http://blog.csdn.net/yeqiuzs/article/details/52166689
