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();
给定一单链表，返回一随机节点的值，每个节点被选中的概率必须一致。
进阶：
如果单链表非常大，且其长度未知的情况将如何？能否在不使用额外空间的情况下解决这个问题？
思路
【Reservoir Sampling 蓄水池抽样问题】
（可理解为为等概抽样问题）
问题：n个数中抽取k个，确保每个数被抽中的概率为n/k。
-
基本思路：
- 先选取1,2,3,...,k将之放入蓄水池；
- 对于k+1，将之以k/(k+1)的概率抽取，然后随机替换水池中的一个数。
- 对于k+i，将之以k/(k+i)的概率抽取，然后随机替换水池中的一个数。
- 重复上述，直到k+i到达n；
证明：
对于k+i，其选中并替换水池中已有元素的概率为k/(k+i)
对于水池中的某数x，其之前就在水池，一次替换后仍在水池中的概率是
P(x之前在水池) * P(未被k+i替换)
=P(x之前在水池) * (1-P(k+i被选中且替换了x) )
= k/(k+i-1) × （1 - k/(k+i) × 1/k）
= k/(k+i)
当k+i到达n，则结果为k/n-
举例
- Choose 3 numbers from [111, 222, 333, 444]. Make sure each number is selected with a probability of 3/4
- First, choose [111, 222, 333] as the initial reservior
- Then choose 444 with a probability of 3/4
- For 111, it stays with a probability of
- P(444 is not selected) + P(444 is selected but it replaces 222 or 333)= 1/4 + 3/4*2/3= 3/4
- The same case with 222 and 333
- Now all the numbers have the probability of 3/4 to be picked
对于本题，取k=1即可
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
private:
ListNode* head;
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;
ListNode* node = head->next;
int i = 2;
while(node){
int j = rand()%i;
if(j==0)
res = node->val;
i++;
node = node->next;
}
return res;
}
};
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(head);
* int param_1 = obj.getRandom();
*/
