382. Linked List Random Node

382. Linked List Random Node

https://leetcode.com/problems/linked-list-random-node/

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.

Approach 1 : Reservoir Sampling

Walk through all .

For each node, if rand() % len == 0 , set the current value to the answer.

Proof:

• For the last node n, the probability is $\frac{1}{n}$ obviously .

  So the probability of left nodes $[0, n-1)$ is $\frac{n-1}{n}$.

• For the n-1 node, the probability is $ \frac{n-1}{n}*\frac{1}{n-1} = \frac{1}{n} $ .

  So the probability of left nodes $[0, n-2)$ is $\frac{n-2}{n}$.

• For the n-2 node. the probability is $ \frac{n-2}{n}*\frac{1}{n-2} = \frac{1}{n} $

  So the probability of left nodes $[0, n-3)$ is $\frac{n-3}{n}$.

• Similarly, we can prove that:

  For ANY node, the probability is $\frac{1}{n}$ .

class Solution {
public:

    ListNode* u;
    Solution(ListNode* head) {
        u = head;
    }
    dd
    int getRandom() {
        int res, len = 1;
        ListNode* v = u;
        while(v){
            if(rand() % len == 0){
                res = v->val;
            }
            len++;
            v = v->next;
        }
        return res;
    }
};

Runtime: 36 ms, faster than 87.72% of C++ online submissions for Linked List Random Node.

Memory Usage: 16.5 MB, less than 50.00% of C++ online submissions for Linked List Random Node.

Complexity Analysis

• Time complexity : $O(n)$
• Space complexity : $O(1)$