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();思路
在遍歷到第i個數時設置選取這個數的概率爲1/i,
對於任意一個數i來說, 其被選擇的概率爲1/i*(1-1/(i+1))…(1-1/n) = 1/n
所以在每一個數的時候我們只要按照隨機一個是否是i的倍數即可決定是否取當前數即可,概率爲 1/i
代碼（C++）
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
ListNode *p;
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() {
int res = 0;
ListNode *temp = p;
for (int cnt = 1; temp != NULL; cnt++, temp = temp->next)
if (rand() % cnt == 0)
res = temp->val;
return res;
}
};
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(head);
* int param_1 = obj.getRandom();
*/
