382. Linked List Random Node

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();

解题思路:

参考了Grandyang的总结

这里使用了水塘抽样法,这位大佬Strugglion 的解释很棒,可以多读读

代码:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
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, i = 2;
        ListNode *cur = head->next;
        while (cur) {
            int j = rand() % i;
            if (j == 0) res = cur->val;
            ++i;
            cur = cur->next;
        }
        return res;
    }
private:
    ListNode *head;
};

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(head);
 * int param_1 = obj.getRandom();
 */
原文地址:https://www.cnblogs.com/yaoyudadudu/p/9237333.html