LeetCode 382. Linked List Random Node

382. 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.

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();

题目大意：给一个单链表，返回一个随机结点的值，要求每一个结点都有同样的概率被选中～
分析：先求出链表的长度len，然后用rand() % len生成一个0～len-1之间的随机数cnt，然后从head开始遍历直到第cnt个结点，返回第cnt个结点的值～

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {private:    ListNode* head;    int len;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;        len = 1;        while(head->next != NULL) {            head = head->next;            len++;        }    }        /** Returns a random node's value. */    int getRandom() {        ListNode* node = head;        int cnt = rand() % len;        while(cnt--) {            node = node->next;        }        return node->val;    }};/** * Your Solution object will be instantiated and called as such: * Solution obj = new Solution(head); * int param_1 = obj.getRandom(); */