【Leetcode链表】反转链表 II（92）

题目

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

解答

一轮指针变换，时间复杂度O(n)，空间复杂度O(1)

# 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.next or m==n:
            return head
        n2, m2 = n, m

        thead = ListNode(-1)
        thead.next = head
        p, p3 = thead, thead
        
        p2 = ListNode(-1)
        p2.next = thead
        c = head

        # 指针就位
        while m:
            p2 = p2.next
            p = p.next
            p3 = p3.next
            c = c.next

            m -= 1
        
        # 反转
        Q = n2-m2
        while Q:
            n = c.next
            c.next = p
            p = c
            c = n

            Q -= 1
        p2.next = p
        p3.next = c

        #　从头反转
        if m2==1 and n2!=m2:
            return p
        return thead.next