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.
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。
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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