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();
题目标签：Array, HashTable, Design
题目让我们设计一个 数据结构， 能插入val， 删除val 和 获得 随机val ，并且是O(1) 时间。
一开始想到的是HashSet，但是对于 取得随机val O(1) 肯定不行。
所以，能在O(1) 时间内 获得随机项，肯定是array。那么这一题需要array 和 HashMap的配合使用。
ArrayList nums 保存所有的val；HashMap 保存 所有的val 和 index（在nums）的映射。
对于insert：nums 直接 add ，map 直接 put，都符合O(1);
对于remove：map remove 符合O(1), 但是 nums remove的话，只有当remove 最后一项的时候，才符合O(1)。如果遇到的val 在nums 里不是最后一项的话，把最后一项的val 保存到 val 的位置，并且要更新最后一项在map中的index，然后nums remove 最后一项。
对于getRandom：直接在nums 里随机返回就可以了。
Java Solution:
Runtime beats 86.91%
完成日期：09/16/2017
关键词：Array, Hash Table, Design
关键点：利用array 保存数值；利用map保存 - 数值 当作key，数值在array里的index 当作value。
class RandomizedSet
{
private HashMap<Integer, Integer> map; // key is value, value is index
private ArrayList<Integer> nums; // store all vals
private java.util.Random rand = new java.util.Random(); /** Initialize your data structure here. */
public RandomizedSet()
{
map = new HashMap<>();
nums = new ArrayList<>();
} /** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
public boolean insert(int val)
{
boolean contain = map.containsKey(val); if(contain)
return false; map.put(val, nums.size());
nums.add(val); return true;
} /** Removes a value from the set. Returns true if the set contained the specified element. */
public boolean remove(int val)
{
boolean contain = map.containsKey(val);
if(!contain)
return false; int valIndex = map.get(val);
if(valIndex != nums.size() - 1) // if this val is not the last one
{
// copy the last one value into this val's position
int lastNum = nums.get(nums.size() - 1);
nums.set(valIndex, lastNum);
// update the lastNum index in map
map.put(lastNum, valIndex);
}
map.remove(val);
nums.remove(nums.size() - 1); // only remove last one is O(1) return true;
} /** Get a random element from the set. */
public int getRandom()
{
return nums.get(rand.nextInt(nums.size()));
}
} /**
* Your RandomizedSet object will be instantiated and called as such:
* RandomizedSet obj = new RandomizedSet();
* boolean param_1 = obj.insert(val);
* boolean param_2 = obj.remove(val);
* int param_3 = obj.getRandom();
*/
参考资料：
https://discuss.leetcode.com/topic/53216/java-solution-using-a-hashmap-and-an-arraylist-along-with-a-follow-up-131-ms
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