876. Middle of the Linked List

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于 2019-03-13 23:49:28 发布

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分类专栏： leetcode

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本文介绍了一种在非空单链表中找到中间节点的方法。如果存在两个中间节点，则返回第二个中间节点。通过快慢指针技巧实现，快指针每次移动两步，慢指针每次移动一步，当快指针到达链表尾部时，慢指针恰好位于中间位置。

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Given a non-empty, singly linked list with head node head, return a middle node of linked list.

If there are two middle nodes, return the second middle node.

Example 1:

Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3. (The judge’s serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def middleNode(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        slow = fast = head
        while fast and fast.next:
			slow = slow.next
			fast = fast.next.next
        return slow