[LeetCode] 382. Linked List Random Node 链表随机节点

Posted 2021-01-10 轻风舞动

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();

 

给一个链表，随机返回一个节点。

解法1：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。但如果n很大就不好处理了。

解法2: 水塘抽样 Reservoir sampling，是一系列的随机算法，其目的在于从包含n个项目的集合S中选取k个样本，其中n为一很大或未知的数量，尤其适用于不能把所有n个项目都存放到内存的情况。

Java:

public class Solution {
    
    ListNode head;
    Random random;
    
    public Solution(ListNode h) {
        head = h;       
        random = new Random();        
    }
    
    public int getRandom() {
        
        ListNode c = head;
        int r = c.val;
        for(int i=1;c.next != null;i++){
            
            c = c.next;
            if(random.nextInt(i + 1) == i) r = c.val;                        
        }
        
        return r;
    }
}　

Python:

from random import randint

class Solution(object):

    def __init__(self, head):
        """
        @param head The linked list‘s head. Note that the head is guanranteed to be not null, so it contains at least one node.
        :type head: ListNode
        """
        self.__head = head


    # Proof of Reservoir Sampling:
    # https://discuss.leetcode.com/topic/53753/brief-explanation-for-reservoir-sampling
    def getRandom(self):
        """
        Returns a random node‘s value.
        :rtype: int
        """
        reservoir = -1
        curr, n = self.__head, 0
        while curr:
            reservoir = curr.val if randint(1, n+1) == 1 else reservoir
            curr, n = curr.next, n+1
        return reservoir

C++: 1

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;
};

C++: 2

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;
};

　　

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