382. Linked List Random Node [medium] (Python)

题目链接

https://leetcode.com/problems/linked-list-random-node/

题目原文

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();

思路方法

思路一

直观的思路：在初始化时统计链表长度，每次随机返回节点时，在该长度内随机出一个索引，从链表头开始找到该索引对应的值即可。空间复杂度O(1)。

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.head = head
        self.length = 0
        while head:
            self.length += 1
            head = head.next

    def getRandom(self):
        """
        Returns a random node's value.
        :rtype: int
        """
        pointer = self.head
        index = random.randint(0, self.length - 1)
        for i in xrange(index):
            pointer = pointer.next
        return pointer.val


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

思路二

如果不考虑follow up的要求，可以将所有元素先保存到数组，那么在随机取数时效率会高的多，但空间复杂度O(n)。

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

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


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

思路三

蓄水池采样（Reservoir Sampling）算法。蓄水池采样解决的是从N个数中等概率的取出k个数的问题，本题特殊情况k=1。

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.head = head

    def getRandom(self):
        """
        Returns a random node's value.
        :rtype: int
        """
        node, n = self.head, 0
        res = 0
        while node:
            if random.randint(0, n) == 0:
                res = node.val
            node, n = node.next, n+1
        return res


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

PS: 写错了或者写的不清楚请帮忙指出，谢谢！
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