leetcode之Linked List Random Node

这题本身来讲是比较简单的。把所有的值放进一个list里，想取哪个取哪个。

import random# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass 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.list1 = []        while head:            self.list1.append(head.val)            head = head.next    def getRandom(self):        """        Returns a random node's value.        :rtype: int        """        location = random.randint(1, len(self.list1))        return self.list1[location - 1]

但是这样是要花费额外时间的，如果按照题目要求的那样，不花费额外的话，做法就相当于每个node都roll个点，roll完一遍之后，谁的点大谁就胜出。