382. Linked List Random Node 链接列表随机节点

Posted 2020-10-22 Long Long Journey

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();

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
import random
class Solution:
    def __init__(self, head):
        """
        @param head The linked list‘s head.
        Note that the head is guaranteed to be not null, so it contains at least one node.
        :type head: ListNode
        """
        self.head = head
        self.length = 0
        node = head
        while node:
            self.length += 1
            node = node.next
    def getRandom(self):
        """
        Returns a random node‘s value.
        :rtype: int
        """
        idx = random.randint(0, self.length - 1)
        node = self.head
        while idx:
            idx -= 1
            node = node.next
        return node.val
# Your Solution object will be instantiated and called as such:
# obj = Solution(head)
# param_1 = obj.getRandom()

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