[leetcode]380. Insert Delete GetRandom O(1)常数时间插入删除取随机值
Design a data structure that supports all following operations in average O(1) time.
insert(val)
: Inserts an item val to the set if not already present.remove(val)
: Removes an item val from the set if present.getRandom
: Returns a random element from current set of elements. Each element must have the same probability of being returned.
Example:
// Init an empty set.
RandomizedSet randomSet = new RandomizedSet(); // Inserts 1 to the set. Returns true as 1 was inserted successfully.
randomSet.insert(1); // Returns false as 2 does not exist in the set.
randomSet.remove(2); // Inserts 2 to the set, returns true. Set now contains [1,2].
randomSet.insert(2); // getRandom should return either 1 or 2 randomly.
randomSet.getRandom(); // Removes 1 from the set, returns true. Set now contains [2].
randomSet.remove(1); // 2 was already in the set, so return false.
randomSet.insert(2); // Since 2 is the only number in the set, getRandom always return 2.
randomSet.getRandom();
Solution1: HashMap + ArrayList
1. use an ArrayList together with HashMap, making hashmap save key(item)->value(idx)
2. To reverse item in O(1), avoiding traversal the whole arrayList in O(n) time, we swap the toDelete item with last item in the list
3. Go through an example like this:
use HashMap to get removeIdx:
Set the last item in the list as replaceValue, why the last item? In order to maintain other indices.
Deal with HashMap: (1) put (2)delete
Deal with List: (1) set (2)delete
code:
class RandomizedSet {
Map<Integer, Integer> map;
List<Integer> list;
Random r; // java 自带类 /** Initialize your data structure here. */
public RandomizedSet() {
list = new ArrayList<>();
map = new HashMap<>();
r = 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) {
/**
hashmap
30 - 0
40 - 1
50 - 2
60 - 3 list: 30 40 50 60
0 1 2 3
**/
if (!map.containsKey(val)) return false;
int removeIdx = map.get(val); // Idx: 1
int replaceValue = list.get(list.size()-1); // replaceValue : 60 // deal with map
map.put(replaceValue, removeIdx);
map.remove(val); // deal with list
list.set(removeIdx, replaceValue);
list.remove(list.size() -1); return true;
} /** Get a random element from the set. */
public int getRandom() {
return list.get(r.nextInt(list.size()));
}
}
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