linked list 经典题

快慢指针找中点(偶数数列时可选择前一个中点还是后一个中点:这里选择前一个中点) 反转列表(反转时要head，mid 会改变) head 从 1->2->3->4->N 变为 1->2->3->N 再反转中3的 next 被从4 改为null

merget'wolists 一定不能忘了左边的list要制定end node None: mid.next = None

 

 

 

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def middleNode(self,head):
        # return the first middle node when array is of even length
        fast,slow = head,head
        while fast and fast.next and fast.next.next:
            fast = fast.next.next
            slow = slow.next
        return slow
    def reverseList(self, start):
        pre,cur = None, start
        while cur:
            temp = cur.next
            cur.next = pre
            pre = cur
            cur = temp
        return pre
    def mergeList(self, l1,l2):
        while l1 and l2:
            l1_temp = l1.next
            l2_temp = l2.next

            l1.next = l2
            l1 = l1_temp

            l2.next = l1
            l2 = l2_temp

    def reorderList(self, head: ListNode) -> None:
        """
        Do not return anything, modify head in-place instead.
        """
        print(head)
        if not head: 
            return None

        mid = self.middleNode(head)
        l2 = self.reverseList(mid.next)
        
        mid.next = None
        self.mergeList(head,l2)