【python3】leetcode 876. Middle of the Linked Listr (easy)
阿新 • • 發佈:2019-01-08
876. Middle of the Linked Listr (easy)
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.
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.
蠢辦法 先遍歷一遍求長度,再遍歷長度的一半
class Solution: def middleNode(self, head): """ :type head: ListNode :rtype: ListNode """ n = 1 node = head while(node.next): n += 1 node = node.next m = int(n/2) while(m > 0): head = head.next m -= 1 return head
記錄solution的一個辦法,使用list儲存每個node
class Solution(object):
def middleNode(self, head):
A = [head]
while(head.next):
head = head.next
A.append(head)
return A[int(len(A)/2)]