题目大意：写一个方法返回列表中的随机元素。
Problem:
https://leetcode.com/problems/linked-list-random-node/description/
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:
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// 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(); |
Solution:
C++
Time Complexity: O(n)
Space Complexity: O(1)
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// Author: Huahua // Running time: 43 ms 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) { head_ = head; n_ = 0; ListNode* curr = head; while (curr != nullptr) { curr = curr->next; ++n_; } } /** Returns a random node's value. */ int getRandom() { int n = rand() % n_; ListNode* curr = head_; while (n-->0) curr = curr->next; return curr->val; } private: int n_; // Does not own the object ListNode* head_; }; |
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