LeetCode 380. Insert Delete GetRandom O(1) (插入删除和获得随机数 常数时间)

Design a data structure that supports all following operations in average O(1) time.

 

  1. insert(val): Inserts an item val to the set if not already present.
  2. remove(val): Removes an item val from the set if present.
  3. 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。

 1 class RandomizedSet 
 2 {
 3     private HashMap<Integer, Integer> map; // key is value, value is index
 4     private ArrayList<Integer> nums; // store all vals
 5     private java.util.Random rand = new java.util.Random();
 6     
 7     /** Initialize your data structure here. */
 8     public RandomizedSet() 
 9     {
10         map = new HashMap<>();
11         nums = new ArrayList<>();
12     }
13     
14     /** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
15     public boolean insert(int val) 
16     {
17         boolean contain = map.containsKey(val);
18         
19         if(contain)
20             return false;
21         
22         map.put(val, nums.size());
23         nums.add(val);
24         
25         return true;
26     }
27     
28     /** Removes a value from the set. Returns true if the set contained the specified element. */
29     public boolean remove(int val) 
30     {
31         boolean contain = map.containsKey(val);
32         if(!contain) 
33             return false;
34         
35         int valIndex = map.get(val);
36         if(valIndex != nums.size() - 1) // if this val is not the last one
37         {
38             // copy the last one value into this val's position
39             int lastNum = nums.get(nums.size() - 1);
40             nums.set(valIndex, lastNum);
41             // update the lastNum index in map
42             map.put(lastNum, valIndex);
43         }
44         map.remove(val);
45         nums.remove(nums.size() - 1); // only remove last one is O(1)
46         
47         return true;
48     }
49     
50     /** Get a random element from the set. */
51     public int getRandom() 
52     {
53         return nums.get(rand.nextInt(nums.size()));
54     }
55 }
56 
57 /**
58  * Your RandomizedSet object will be instantiated and called as such:
59  * RandomizedSet obj = new RandomizedSet();
60  * boolean param_1 = obj.insert(val);
61  * boolean param_2 = obj.remove(val);
62  * int param_3 = obj.getRandom();
63  */

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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posted @ 2017-09-17 10:52  Jimmy_Cheng  阅读(360)  评论(0编辑  收藏  举报