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社区首页 >专栏 >LeetCode 0382 - Linked List Random Node

LeetCode 0382 - Linked List Random Node

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Reck Zhang
发布2021-08-11 11:43:15
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发布2021-08-11 11:43:15
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文章被收录于专栏:Reck Zhang

Linked List Random Node

Desicription

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:

代码语言:javascript
复制
// 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

代码语言:javascript
复制
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
private:
    ListNode* head = nullptr;
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->head = head;
    }

    /** Returns a random node's value. */
    int getRandom() {
        auto result = head->val;
        auto cur = head->next;
        int i = 2;

        while(cur != nullptr) {
            if(rand() % i == 0) {
                result = cur->val;
            }
            i++;
            cur = cur->next;
        }

        return result;
    }
};

/**
 * Your Solution object will be instantiated and called as such:
 * Solution* obj = new Solution(head);
 * int param_1 = obj->getRandom();
 */
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目录
  • Linked List Random Node
    • Desicription
      • Solution
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