382 Linked List Random Node.py

"""
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();
Author: Rajeev Ranjan

"""

import random


# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution(object):
    def __init__(self, head):
        """
        Reservoir sampling
        :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

    def getRandom(self):
        """
        Returns a random node's value.
        :rtype: int
        """
        ret = self.head
        cur = self.head.next
        idx = 1
        while cur:
            if random.randrange(0, idx+1) == 0:
                ret = cur
            cur = cur.next
            idx += 1

        return ret.val


# Your Solution object will be instantiated and called as such:
# obj = Solution(head)
# param_1 = obj.getRandom()