［抄题］：
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();
[暴力解法]：
时间分析：
空间分析：
[优化后]：
时间分析：
空间分析：
［奇葩输出条件］：
［奇葩corner case］：
如果hashmap的value为空，则必须把key也删掉
［思维问题］：
［一句话思路］：
用set存location, 然后还是先移动 后删除
［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：
［画图］：
［一刷］：
- 提前判断存在性还是需要的，因为如果不存在时，需要初始化一个hashset
- map中存的是nums数组的大小，表示目前元素插入成为了新数组最后一位：nums.size()
- nums.remove(index，不是内容)
［二刷］：
［三刷］：
［四刷］：
［五刷］：
[五分钟肉眼debug的结果]：
［总结］：
hashset可以分区
［复杂度］：Time complexity: O() Space complexity: O()
［英文数据结构或算法，为什么不用别的数据结构或算法］：
［算法思想：递归/分治/贪心］：
［关键模板化代码］：
［其他解法］：
［Follow Up］：
［LC给出的题目变变变］：
[代码风格] ：
class RandomizedCollection {
/** Initialize your data structure here. */
HashMap<Integer, Set<Integer>> map;
ArrayList<Integer> nums;
java.util.Random rand = new java.util.Random();
public RandomizedCollection() {
map = new HashMap<Integer, Set<Integer>>();
nums = new ArrayList<>();
}
/** Inserts a value to the collection. Returns true if the collection did not already contain the specified element. */
public boolean insert(int val) {
boolean contain = map.containsKey(val);
if (!contain) map.put(val, new HashSet<Integer>());
map.get(val).add(nums.size());
nums.add(val);
return !contain;
}
/** Removes a value from the collection. Returns true if the collection contained the specified element. */
public boolean remove(int val) {
boolean contain = map.containsKey(val);
if (!contain) return false;
int loc = map.get(val).iterator().next();
//renew location
if (loc < nums.size() - 1) {
int lastone = nums.get(nums.size() - 1);
nums.set(loc, lastone);
map.get(lastone).add(loc);
map.get(lastone).remove(nums.size() - 1);
}
//remove, if key is empty
nums.remove(nums.size() - 1);
map.get(val).remove(loc);
if (map.get(val).isEmpty()) map.remove(val);
//return
return true;
}
/** Get a random element from the collection. */
public int getRandom() {
return nums.get(rand.nextInt(nums.size()));
}
}
/**
* Your RandomizedCollection object will be instantiated and called as such:
* RandomizedCollection obj = new RandomizedCollection();
* boolean param_1 = obj.insert(val);
* boolean param_2 = obj.remove(val);
* int param_3 = obj.getRandom();
*/