Reservoir Sampling-382. Linked List Random Node
阿新 • • 發佈:2018-01-13
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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();
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):cur(head) { } /** Returns a random node‘s value. */ int getRandom() { int val=cur->val; ListNode*temp=cur; for(int i=0;temp!=nullptr;temp=temp->next,++i) { uniform_int_distribution<unsigned> u(0,i); default_random_engine e(rand());//真正隨機的種子 unsigned int m=u(e); if(m<1) { val=temp->val; } }return val; } private: ListNode *cur; };
Reservoir Sampling-382. Linked List Random Node