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  • Reservoir Sampling-382. Linked List Random Node

    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());//真正随机的种子
                unsigned int m=u(e);
                if(m<1)
                {
                    val=temp->val;
                }
            }
            return val;
        }
    private:
        ListNode *cur;
    };
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  • 原文地址:https://www.cnblogs.com/msymm/p/8278258.html
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