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  • 0x0 题目详情
  • 0x1 解题思路
  • 0x2 代码实现
  • 0x3 课后总结

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  1. hash-table

380-Insert-Delete-GetRandom-O(1)

Previous442-Find-All-Duplicates-in-an-ArrayNext1-Two-Sum

Last updated 4 years ago

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0x0 题目详情

设计一个支持在平均 时间复杂度 O(1) 下,执行以下操作的数据结构。 insert(val):当元素 val 不存在时,向集合中插入该项。 remove(val):元素 val 存在时,从集合中移除该项。 getRandom:随机返回现有集合中的一项。每个元素应该有相同的概率被返回。

测试用例:

示例 : // 初始化一个空的集合。 RandomizedSet randomSet = new RandomizedSet(); // 向集合中插入 1 。返回 true 表示 1 被成功地插入。 randomSet.insert(1); // 返回 false ,表示集合中不存在 2 。 randomSet.remove(2); // 向集合中插入 2 。返回 true 。集合现在包含 [1,2] 。 randomSet.insert(2); // getRandom 应随机返回 1 或 2 。 randomSet.getRandom(); // 从集合中移除 1 ,返回 true 。集合现在包含 [2] 。 randomSet.remove(1); // 2 已在集合中,所以返回 false 。 randomSet.insert(2); // 由于 2 是集合中唯一的数字,getRandom 总是返回 2 。 randomSet.getRandom();

0x1 解题思路

这道题需要通过维护两个哈希表实现。一张是元素和未插入时对应size,一张是size大小和其对应的元素。下标就是当前插入的位置。为size。

class RandomizedSet {
    Map<Integer,Integer> keyIndexMap;
    Map<Integer,Integer> indexKeyMap;
    int size;
....
}

当插入元素时,更新俩张哈希表,并更新size。删除元素时,因为我们要随机选取元素,不同的size对应不同的元素,一旦中间的元素被移除了,在0~size-1范围内就会出现size存在,对应元素不存在的情况。所以为了避免这种情况,我们每删除一个元素后,就拿最后一个元素填充到被删除的位置,并更新size,这样就满足了随机获取的元素的要求。

0x2 代码实现

class RandomizedSet {
    Map<Integer,Integer> keyIndexMap;
    Map<Integer,Integer> indexKeyMap;
    int size;

    /** Initialize your data structure here. */
    public RandomizedSet() {
        keyIndexMap=new HashMap<>();
        indexKeyMap=new HashMap<>();
        size=0;
    }

    /** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
    public boolean insert(int val) {
        if(keyIndexMap.containsKey(val)){
            return false;
        }
        //一个size对应一个元素
        keyIndexMap.put(val,size);
        indexKeyMap.put(size,val);
        size++;
        return true;
    }

    /** Removes a value from the set. Returns true if the set contained the specified element. */
    public boolean remove(int val) {
        if(!keyIndexMap.containsKey(val)){
            return false;
        }
        int deleteIndex=keyIndexMap.get(val);
        int lastKey=indexKeyMap.get(size-1);
        //将最后一个位置上的元素填充被删除的位置
        keyIndexMap.put(replaceKey,deleteIndex);
        indexKeyMap.put(deleteIndex,lastKey);
        keyIndexMap.remove(val);
        indexKeyMap.remove(size-1);
        size--;
        return true;
    }

    /** Get a random element from the set. */
    public int getRandom() {
        if(this.size==0){
            return -1;
        }
        //获取的随机数在[0,size)范围内
        int random=(int)(Math.random()*(this.size));
        return this.indexKeyMap.get(random);
    }
}

0x3 课后总结

答案是抄的,没什么好说的。

原题链接