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();
最暴力的方法:
1)构建一个数组,存下每一个node的地址,随机取一个。
空间复杂度O(n),时间复杂度O(2n)
次暴力方法的方法:
1)遍历一遍,求长度n
2) 再从头遍历,第i个节点的取中概率为。取中即返回。
证明,遍历到第i个节点时,返回的概率为
空间复杂度O(1),时间复杂度O(2n)
较好方法:
1)建立一个临时变量result
2)遍历链表,第i个节点以概率把值赋给result。
3)遍历完成后返回result。
result中存有第i个节点的值的概率为:
空间复杂度O(1),时间复杂度O(n)
code:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
#define random(x) (rand()%x)
class Solution {
private:
ListNode *list;
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) {
this->list = head;
}
/** Returns a random node's value. */
int getRandom() {
ListNode *p = this->list->next;
int result = this->list->val;
for(size_t i=2; p != nullptr; i++){
if(random(i) == 0){
result = p->val;
}
p = p->next;
}
return result;
}
};
/**
* Your Solution object will be instantiated and called as such:
* Solution* obj = new Solution(head);
* int param_1 = obj->getRandom();
*/