Linked List Random Node

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本文介绍了一种算法，用于从链表中随机选择一个节点，确保每个节点被选中的概率相同。即使链表非常大且长度未知，该算法也能高效运行且不使用额外的空间。

一、问题描述

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.

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?

Example:

// Init a singly linked list [1,2,3].
ListNode head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
Solution solution = new Solution(head);

// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.
solution.getRandom();

二、思路

问题描述：返回链表的一个随机的值。

三、代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
private:
     ListNode* head;   
    
public:
    /** @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node. */
 
    Solution(ListNode* head) {
        this -> head = head;
    }
    /** Returns a random node's value. */
    int getRandom() {
        int res = head -> val;
        ListNode* pNode = head -> next;
        int i = 2;
        while(pNode){
            int j = rand() % i;
            if(j == 0) res = pNode -> val;
            ++i;
            pNode = pNode -> next;
        }
        return res;
    }
};

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(head);
 * int param_1 = obj.getRandom();
 */