title
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();
solution
这道题让我们在常数时间范围内实现插入删除和获得随机数操作，如果这道题没有常数时间的限制，那么将会是一道非常简单的题，我们直接用一个set就可以搞定所有的操作。但是由于时间的限制，我们无法在常数时间内实现获取随机数，所以只能另辟蹊径。此题的正确解法是利用到了一个一维数组和一个哈希表，其中数组用来保存数字，哈希表用来建立每个数字和其在数组中的位置之间的映射，对于插入操作，我们先看这个数字是否已经在哈希表中存在，如果存在的话直接返回false，不存在的话，我们将其插入到数组的末尾，然后建立数字和其位置的映射。删除操作是比较tricky的，我们还是要先判断其是否在哈希表里，如果没有，直接返回false。由于哈希表的删除是常数时间的，而数组并不是，为了使数组删除也能常数级，我们实际上将要删除的数字和数组的最后一个数字调换个位置，然后修改对应的哈希表中的值，这样我们只需要删除数组的最后一个元素即可，保证了常数时间内的删除。而返回随机数对于数组来说就很简单了，我们只要随机生成一个位置，返回该位置上的数字即可
class RandomizedSet {
private:
unordered_map<int, int> map1;
vector<int> v;
public:
/** Initialize your data structure here. */
RandomizedSet() {
}
/** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
bool insert(int val) {
if (map1.count(val)) {
return false;
}
v.push_back(val);
map1[val] = v.size() - 1;
return true;
}
/** Removes a value from the set. Returns true if the set contained the specified element. */
bool remove(int val) {
if (!map1.count(val)) {
return false;
}
map1[v.back()] = map1[val];
v[map1[val]] = v.back();
v.pop_back();
map1.erase(val);
return true;
}
/** Get a random element from the set. */
int getRandom() {
return v[rand() % v.size()];
}
};
Java solution with two HashMaps: easy to understand
import java.util.Random;
public class RandomizedSet {
private HashMap<Integer, Integer> keyMap = null;
private HashMap<Integer, Integer> valueMap = null;
int count;
/** Initialize your data structure here. */
public RandomizedSet() {
keyMap = new HashMap<Integer, Integer>();
valueMap = new HashMap<Integer, Integer>();
}
/** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
public boolean insert(int val) {
if(keyMap.containsKey(val)) {
return false;
} else {
keyMap.put(val, count);
valueMap.put(count, val);
count = keyMap.size();
return true;
}
}
/** Removes a value from the set. Returns true if the set contained the specified element. */
public boolean remove(int val) {
if(!keyMap.containsKey(val)) {
return false;
} else {
int valueKey = keyMap.get(val);
keyMap.remove(val);
if(valueKey != valueMap.size() - 1) {
valueMap.put(valueKey, valueMap.get(valueMap.size() - 1));
keyMap.put(valueMap.get(valueMap.size() - 1), valueKey);
valueMap.remove(valueMap.size() - 1);
} else {
valueMap.remove(valueKey);
}
count = keyMap.size();
return true;
}
}
/** Get a random element from the set. */
public int getRandom() {
Random random = new Random();
int n = random.nextInt(keyMap.size());
return valueMap.get(n);
}
}