【LeetCode】382. Linked List Random Node 解题报告（Python）

【LeetCode】382. Linked List Random Node 解题报告（Python）

标签： LeetCode

题目地址：https://leetcode.com/problems/linked-list-random-node/description/

题目描述：

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();

题目大意

随机从链表中抽出一个节点的数字。

解题方法

我使用一个数组保存了，然后从中间随机找的index。

代码：

# 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.stack = []
while head:
self.stack.append(head.val)
head = head.next

def getRandom(self):
"""
Returns a random node's value.
:rtype: int
"""
_len = len(self.stack)
return self.stack[random.randint(0, _len - 1)]

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

方法二：

蓄水池抽样。。做法待补。

日期

2018 年 3 月 8 日