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  • 92. Reverse Linked List II

    1. 原始题目

    Reverse a linked list from position m to n. Do it in one-pass.

    Note: 1 ≤ m ≤ n ≤ length of list.

    Example:

    Input: 1->2->3->4->5->NULL, m = 2, n = 4
    Output: 1->4->3->2->5->NULL

    2. 题目理解

    将m至n的链表反转。注意下标这里在链表中实际上是m-1至n-1。

    我的思路是先遍历m步,然后在n-m+1遍历过程中将结点反转。然后将这一整段链表重新链接即可。这样做一个缺点就是需要多个标记点。

    3. 解题

     1 # Definition for singly-linked list.
     2 # class ListNode:
     3 #     def __init__(self, x):
     4 #         self.val = x
     5 #         self.next = None
     6 
     7 class Solution:
     8     def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
     9         if not head.next:
    10             return head
    11         m = m-1                  # 修正下标
    12         n = n-1
    13         dummy = ListNode(0)
    14         dummy.next = head
    15         record = dummy
    16         
    17         for _ in range(m):       # 先走到需要反转的地方
    18             dummy = dummy.next
    19             
    20         end = dummy.next         # 记录断点
    21         i = dummy
    22         j = i.next               # i,j,k 负责反转链表
    23         if j:
    24             k = j.next
    25             for _ in range(n-m+1):    
    26                 if j:
    27                     j.next = i
    28                     i = j
    29                     j = k
    30                     if k:
    31                         k = k.next
    32         else:
    33             return head
    34         dummy.next = i          # 重新链接链表一端
    35         end.next = j            # 重新连接链表另一端
    36             
    37         return record.next

    验证:

     1 # 新建链表1
     2 listnode1 = ListNode_handle(None)
     3 s1 = [1,2,3,44,555,66,7,8]
     4 for i in s1:
     5     listnode1.add(i)
     6 
     7 listnode1.print_node(listnode1.head)
     8 
     9 s = Solution()
    10 head = s.reverseBetween(listnode1.head,4,5)
    11 listnode1.print_node(head)

    1 2 3 44 555 66 7 8
    1 2 3 555 44 66 7 8

    LeetCode上有另一种解题思路和我的极其类似:这里粘过来对比学习一下:

     1 class Solution(object):
     2     def reverseBetween(self, head, m, n):
     3         """
     4         :type head: ListNode
     5         :type m: int
     6         :type n: int
     7         :rtype: ListNode
     8         """
     9         if not head or m == n: return head
    10         p = dummy = ListNode(None)
    11         dummy.next = head
    12         for i in range(m-1): p = p.next
    13         tail = p.next
    14 
    15         for i in range(n-m):
    16             tmp = p.next                  # a)
    17             p.next = tail.next            # b)
    18             tail.next = tail.next.next    # c)
    19             p.next.next = tmp             # d)
    20         return dummy.next

    流程如下:

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  • 原文地址:https://www.cnblogs.com/king-lps/p/10664376.html
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