linked-list-random-node.py

# Time:  O(n)
# Space: O(1)

# 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();


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


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