题解 | #牛群的重新排列#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @param left int整型 
# @param right int整型 
# @return ListNode类
#
class Solution:
    def reverseBetween(self , head: ListNode, left: int, right: int) -> ListNode:
        dummy = ListNode(0)
        dummy.next = head
        pre = dummy

        for i in range(left - 1):
            pre = pre.next
  
        cur = pre.next
        for i in range(right - left):
            nxt = cur.next
            cur.next = nxt.next
            nxt.next = pre.next
            pre.next = nxt

        return dummy.next

• 题目考察的知识点 : 链表的区间反转
• 题目解答方法的文字分析：

1. 定义dummy节点指向头部，方便头部操作
2. 移动pre指针到反转的起点前面
3. 定义cur指针指向反转区间起点
4. 进行区间反转,依次取nxt,改变指针方向
5. 返回dummy.next即可

• 本题解析所用的编程语言：Python
• 完整且正确的编程代码

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @param left int整型 
# @param right int整型 
# @return ListNode类
#
class Solution:
    def reverseBetween(self , head: ListNode, left: int, right: int) -> ListNode:
        dummy = ListNode(0)
        dummy.next = head
        pre = dummy

        for i in range(left - 1):
            pre = pre.next
  
        cur = pre.next
        for i in range(right - left):
            nxt = cur.next
            cur.next = nxt.next
            nxt.next = pre.next
            pre.next = nxt

        return dummy.next