[LeetCode] 382. Linked List Random Node 链表随机结点

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.

Implement the Solution class:

Solution(ListNode head) Initializes the object with the integer array nums.
int getRandom() Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be choosen.

Example 1:

Input
["Solution", "getRandom", "getRandom", "getRandom", "getRandom", "getRandom"]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 3, 2, 2, 3]

Explanation
Solution solution = new Solution([1, 2, 3]);
solution.getRandom(); // return 1
solution.getRandom(); // return 3
solution.getRandom(); // return 2
solution.getRandom(); // return 2
solution.getRandom(); // return 3
// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.

Constraints:

The number of nodes in the linked list will be in the range [1, 104].
-104 <= Node.val <= 104
At most 104 calls will be made to getRandom.

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?

这道题给了一个链表，让随机返回一个结点，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:
    Solution(ListNode* head) {
        len = 0;
        ListNode *cur = head;
        this->head = head;
        while (cur) {
            ++len;
            cur = cur->next;
        }
    }
    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，没法提前知道长度，这里用到了著名了水塘抽样 Reservoir Sampling 的思路，由于限定了 head 一定存在，所以先让返回值 res 等于 head 的结点值，然后让 cur 指向 head 的下一个结点，定义一个变量i，初始化为2，若 cur 不为空则开始循环，在 [0, i - 1] 中取一个随机数，如果取出来0，则更新 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，这里其实相当于维护了一个大小为1的水塘，然后随机数生成为0的话，交换水塘中的值和当前遍历到的值，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:
    Solution(ListNode* head) {
        this->head = head;
    }
    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;
};