382. Linked List Random Node

应用reservior sampling ，第i个被选中的概率为1/i，之后没有被选中的概率i－1/i，所以第1个被选中的概率为1/1*（1/2）＊（2/3）＊...(n-1)/n=1/n，

class Solution {
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) {
p=head;
}

/** Returns a random node's value. */
int getRandom() {
ListNode* h=p;
int val=h->val;
int len=0;
while(h)
{
if(rand()%(++len)==0)
val=h->val;
h=h->next;
}
return val;
}
private:
ListNode* p;
};