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  • LeetCode 381. Insert Delete GetRandom O(1)

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

    Note: Duplicate elements are allowed.
    1. insert(val): Inserts an item val to the collection.
    2. remove(val): Removes an item val from the collection if present.
    3. getRandom: Returns a random element from current collection of elements. The probability of each element being returned is linearly related to the number of same value the collection contains.

    Example:

    // Init an empty collection.
    RandomizedCollection collection = new RandomizedCollection();
    
    // Inserts 1 to the collection. Returns true as the collection did not contain 1.
    collection.insert(1);
    
    // Inserts another 1 to the collection. Returns false as the collection contained 1. Collection now contains [1,1].
    collection.insert(1);
    
    // Inserts 2 to the collection, returns true. Collection now contains [1,1,2].
    collection.insert(2);
    
    // getRandom should return 1 with the probability 2/3, and returns 2 with the probability 1/3.
    collection.getRandom();
    
    // Removes 1 from the collection, returns true. Collection now contains [1,2].
    collection.remove(1);
    
    // getRandom should return 1 and 2 both equally likely.
    collection.getRandom();
    

    题目标签:Array, Hash Table, Design
      这道题目基于#380 的情况下,可以允许有重复项。回顾一下380, 因为要达到 insert, remove 和 getRandom 都是O(1) 的时间,我们需要ArrayList 来保存所有的数字,利用map 来保存 数字 和 index 之间的映射。
      这道题目允许了重复项,那么在map 里 key 是数字, value 是index, 这里的index 就会不止一个了。我们要把所有重复的数字的 index 也保存进来, 所以把 map 里value 改成 HashSet 来保存所有的index。
     
      insert val:如果map里没有val,需要新建一个HashSet,并且加入nums (ArrayList);
             如果有val,直接加入新的index 进HashSet,并且加入nums (ArrayList);  
            
      remove val:如果val 在nums 里是最后一个的话,只需要在map 里删除val 的index, 并且在nums 里删除最后一个数字。
            如果val 在nums 里不是最后一个的话,需要额外的把 nums 里最后一个数字的值 复制到 val, 删除val 在map里的 index,还要把最后一个数字的index 在map 里更新,并且在nums 里删除最后一个数字。
     
      其他基本都和#380 差不多,具体看code。
     

    Java Solution:

    Runtime beats 88.45% 

    完成日期:09/18/2017

    关键词:Array, Hash Table, Design

    关键点:利用array 保存数值;利用map<Integer, HashSet<>>保存 - 数值 当作key,数值在array里的所有index 保存在HashSet,当作value。

     1 class RandomizedCollection 
     2 {
     3     private HashMap<Integer, HashSet<Integer>> map; // key is value, value is index HashSet
     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 RandomizedCollection() 
     9     {
    10         map = new HashMap<>();
    11         nums = new ArrayList<>();
    12     }
    13     
    14     /** Inserts a value to the collection. Returns true if the collection did not already contain the specified element. */
    15     public boolean insert(int val) 
    16     {
    17         boolean contain = map.containsKey(val);
    18         
    19         // if map doesn't have val, meaning map doesn't have HashSet
    20         if(!contain) 
    21             map.put(val, new HashSet<Integer>()); // create HashSet
    22         
    23         // add index into HashSet
    24         map.get(val).add(nums.size());
    25         nums.add(val);
    26         
    27         return !contain; // if collection has val, return false; else return true
    28     }
    29     
    30     /** Removes a value from the collection. Returns true if the collection contained the specified element. */
    31     public boolean remove(int val) 
    32     {
    33         boolean contain = map.containsKey(val);
    34         if(!contain) 
    35             return false;
    36         // get an index from HashSet of Map
    37         int valIndex = map.get(val).iterator().next();
    38         map.get(val).remove(valIndex); // remove this index from val's set
    39         
    40         if(valIndex != nums.size() - 1) // if this val is not the last one in nums
    41         {
    42             // copy the last one value into this val's position
    43             int lastNum = nums.get(nums.size() - 1);
    44             nums.set(valIndex, lastNum);
    45             // update the lastNum index in HashSet
    46             map.get(lastNum).remove(nums.size() - 1); // remove the last number's index from set
    47             map.get(lastNum).add(valIndex); // add new index into set
    48         }
    49         
    50         if(map.get(val).isEmpty()) // if val's set is empty
    51             map.remove(val); // remove val from map
    52         
    53         nums.remove(nums.size() - 1); // only remove last one O(1)
    54           
    55         return true;
    56     }
    57     
    58     /** Get a random element from the collection. */
    59     public int getRandom() 
    60     {
    61         return nums.get(rand.nextInt(nums.size()));
    62     }
    63 }
    64 
    65 /**
    66  * Your RandomizedCollection object will be instantiated and called as such:
    67  * RandomizedCollection obj = new RandomizedCollection();
    68  * boolean param_1 = obj.insert(val);
    69  * boolean param_2 = obj.remove(val);
    70  * int param_3 = obj.getRandom();
    71  */

    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/5

    LeetCode 题目列表 - LeetCode Questions List
     
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  • 原文地址:https://www.cnblogs.com/jimmycheng/p/7555771.html
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