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();
这道题给了我们一个链表，让我们随机返回一个节点，那么最直接的方法就是先统计出链表的长度，然后根据长度随机生成一个位置，然后从开头遍历到这个位置即可，参见代码如下：
解法一：
class Solution {public: /** @param head The linked list's head. Note that the head is guanranteed to be not null, so it contains at least one node. */ Solution(ListNode* head) { len = 0; ListNode *cur = head; this->head = head; while (cur) { ++len; cur = cur->next; } } /** Returns a random node's value. */ int getRandom() { int t = rand() % len; ListNode *cur = head; while (t) { --t; cur = cur->next; } return cur->val; }private: int len; ListNode *head;}; Follow up中说链表可能很长，我们没法提前知道长度，这里用到了著名了的思路，由于限定了head一定存在，所以我们先让返回值res等于head的节点值，然后让cur指向head的下一个节点，定义一个变量i，初始化为2，若cur不为空我们开始循环，我们在[0, i - 1]中取一个随机数，如果取出来0，那么我们更新res为当前的cur的节点值，然后此时i自增一，cur指向其下一个位置，这里其实相当于我们维护了一个大小为1的水塘，然后我们随机数生成为0的话，我们交换水塘中的值和当前遍历到底值，这样可以保证每个数字的概率相等，参见代码如下：
解法二：
class Solution {public: /** @param head The linked list's head. Note that the head is guanranteed 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;}; 本文转自博客园Grandyang的博客，原文链接：，如需转载请自行联系原博主。