2018-11-09

leetcode

876. Middle of the Linked List

Given a non-empty, singly linked list with head nodehead, 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.

Example 2:

Input: [1,2,3,4,5,6]Output: Node 4 from this list (Serialization:[4,5,6])Since the list has two middle nodes with values 3 and 4, we return the second one.

Note:

The number of nodes in the given list will be between1and100.

分析：当斌看到这道题目后觉得这简直不是leetcode上面的因为太简单了。。。。。。，我仅仅只是遍历统计了一下节点总数，然后取到其中的节点一半的时候就返回节点，然后就解决了。。。。。。尴尬癌都犯了。。。。。。

# 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

        """

        cur = head

        count = 0

        i = 0

        while cur!= None:

            count+=1

            cur = cur.next

        print(count)

        cur = head

        if count == 0:

            return None

        n = count//2

        while i

            cur = cur.next

            i+=1

        return cur