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();1
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35# 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):
"""
@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.xlist = []
node = head
while node:
self.xlist.append(node.val)
if node.next:
node = node.next
else:break
def getRandom(self):
"""
Returns a random node's value.
:rtype: int
"""
import random
return self.xlist[random.randint(0,len(self.xlist)-1)]
# Your Solution object will be instantiated and called as such:
# obj = Solution(head)
# param_1 = obj.getRandom()