LeetCode:143. 重排链表

Posted 2021-06-27 南岸青栀*

143. 重排链表

法1：借助列表存储节点对象

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reorderList(self, head: ListNode) -> None:
        """
        Do not return anything, modify head in-place instead.
        """
        #借用列表，实现重排链表
        lst = []
        p = head
        while p:
            lst.append(p)
            p = p.next
        p = head
        print((len(lst))//2)
        for i in range((len(lst)-1)//2):
            lst[len(lst)-i-1].next = lst[i+1]
            lst[i].next = lst[len(lst)-1-i]
            lst[len(lst)-i-2].next = None

法2：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reorderList(self, head: ListNode) -> None:
        """
        Do not return anything, modify head in-place instead.
        """
        #找到中点
        pre,slow = head,head
        while pre and pre.next:
            pre = pre.next.next
            slow = slow.next
        mid = slow.next
        slow.next = None

        #反转后半段
        head2 = None
        while mid:
            temp = mid.next
            mid.next = head2
            head2,mid = mid,temp
        
        #拼接
        while head2:
            tmp1 = head.next
            tmp2 = head2.next
            head.next = head2
            head2.next = tmp1
            head,head2 = tmp1,tmp2