LeetCode-python 92.反转链表 II

题目链接
难度：中等       类型： 链表

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反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。

说明:
1 ≤ m ≤ n ≤ 链表长度。

示例

输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL

解题思路

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0.保存头节点
1.找到反转部分的前一个节点，保存为start
2.翻转第m到n位链表，记录第m个节点为node_m，第n个节点为node_n，第n+1个节点为end
3.连接链表，start.next = node_n, node_m.next = end
4.返回头节点

代码实现

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
        if not head or not head.next or m==n:
            return head
        dummy = ListNode(0)
        dummy.next = head
        start = dummy
        for i in range(m-1):
            start = start.next
                        
        end = cur = start.next  
        pre = None
        for i in range(n-m+1):
            next = cur.next
            cur.next = pre
            pre = cur
            cur = next            
        start.next = pre
        end.next = cur
        return dummy.next

本文链接：https://www.jianshu.com/p/91b050e076b4