382.Linked List Random Node

tags: Medium,Linked List,Math

382. Linked List Random Node

題目描述

Given a singly linked list, return a random node's value from the linked list. Each node must have the same probability of being chosen.

Implement the Solution class:

• Solution(ListNode head) Initializes the object with the head of the singly-linked list head.
• int getRandom() Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be chosen.

範例

Example 1:

Input
["Solution", "getRandom", "getRandom", "getRandom", "getRandom", "getRandom"]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 3, 2, 2, 3]

Explanation
Solution solution = new Solution([1, 2, 3]);
solution.getRandom(); // return 1
solution.getRandom(); // return 3
solution.getRandom(); // return 2
solution.getRandom(); // return 2
solution.getRandom(); // return 3
// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.

Constraints:

• The number of nodes in the linked list will be in the range [1, 10⁴].
• -10⁴ <= Node.val <= 10⁴
• At most 10⁴ calls will be made to getRandom.

Follow up:

• What if the linked list is extremely large and its length is unknown to you?
• Could you solve this efficiently without using extra space?

解答

C++

思路:

第一種: 先遍歷一次得整體數量n再以此為區間隨機取得數字
第二種: 對於無法先遍歷得整體數量n如streaming data需用Reservoir sampling

class Solution {
public:
    ListNode *head;
    Solution(ListNode* head) {
        this->head = head;
    }
    
    int getRandom() {
        int n = 1, res = 0;
        ListNode* cur = head;
        while(cur)
        {
            int j = rand() % n;
            if(j == 0)
                res = cur->val;
            n++;
            cur = cur->next;
        }
        return res;
    }
};

Time:

$O(n)$
Extra Space:

$O(1)$

XD March 10, 2023

Follow up:

1. 如果選取k個如何做?
2. 如果有m台機器做分散式處理, 全部加總共n個, 選取k個如何做?

Reference

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