LeetCode 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();

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本题要从一个链表中随机取一个数。常规解法是先遍历一遍链表，得到链表长度n，然后rand()%n确定随机取的节点位置，最后遍历链表找到第rand()%n个数。 但是这种解法肯定不是大家想要的，Follow up里提到，如果链表很长，没法提前知道链表长度，该怎么办呢？查阅题解才发现需要用到一个叫做水塘抽样的方法，这篇博客介绍得挺详细的。简单来说，就是假设有n个节点（n提前不知道），我们希望每个节点被采样的概率都是1/n，我们可以这样做，遍历到第i个节点时，以概率1/i采样i号节点。我们来算一算第i号节点被采样的概率=遍历到第i号节点时，要被采样，且后面节点都不被采样，公式为： $$!\frac{1}{i}\ast \frac{i}{i+1}\ast \frac{i+1}{i+2}\ast \dots \ast \frac{n-1}{n}=\frac{1}{n}$$ 即每个节点被采样到的概率都等于$$\frac{1}{n}$$。 具体怎样实现以$$\frac{1}{i}$$采样呢，我们可以令k=rand()%i，则k随机分布在[0,i-1]，所以k==0的概率就是$$\frac{1}{i}$$。完整代码如下： [cpp] class Solution { private: ListNode* mhead; 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) { mhead = head; } /** Returns a random node’s value. */ int getRandom() { int ans = mhead->val, n = 2; ListNode* cur = mhead->next; while (cur) { int r = rand() % n; if (r == 0)ans = cur->val; cur = cur->next; ++n; } return ans; } }; [/cpp] 本代码提交AC，用时49MS。]]>