[LeetCode] Linked List Random Node 链表随机节点

2023年7月28日 251次阅读 来源: Grandyang

 

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();

 

这道题给了我们一个链表，让我们随机返回一个节点，那么最直接的方法就是先统计出链表的长度，然后根据长度随机生成一个位置，然后从开头遍历到这个位置即可，参见代码如下：

 

解法一：

class Solution {
public:
    /** @param head The linked list's head. Note that the head is guanranteed to be not null, so it contains at least one node. */
    Solution(ListNode* head) {
        len = 0;
        ListNode *cur = head;
        this->head = head;
        while (cur) {
            ++len;
            cur = cur->next;
        }
    }
    
    /** Returns a random node's value. */
    int getRandom() {
        int t = rand() % len;
        ListNode *cur = head;
        while (t) {
            --t;
            cur = cur->next;
        }
        return cur->val;
    }
private:
    int len;
    ListNode *head;
};

 

Follow up中说链表可能很长，我们没法提前知道长度，这里用到了著名了水塘抽样Reservoir Sampling的思路，由于限定了head一定存在，所以我们先让返回值res等于head的节点值，然后让cur指向head的下一个节点，定义一个变量i，初始化为2，若cur不为空我们开始循环，我们在[0, i – 1]中取一个随机数，如果取出来0，那么我们更新res为当前的cur的节点值，然后此时i自增一，cur指向其下一个位置，这里其实相当于我们维护了一个大小为1的水塘，然后我们随机数生成为0的话，我们交换水塘中的值和当前遍历到的值，这样可以保证每个数字的概率相等，参见代码如下：

 

解法二：

class Solution {
public:
    /** @param head The linked list's head. Note that the head is guanranteed 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() {
        int res = head->val, i = 2;
        ListNode *cur = head->next;
        while (cur) {
            int j = rand() % i;
            if (j == 0) res = cur->val;
            ++i;
            cur = cur->next;
        }
        return res;
    }
private:
    ListNode *head;
};

 

参考资料：

https://discuss.leetcode.com/topic/53812/using-reservoir-sampling-o-1-space-o-n-time-complexity-c

 

    原文作者：Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/5759926.html
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。
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