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

class Solution:
    def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
        pre = None
        cur = head
        i = 0
        p1 = p2 = p3 = p4 = None
        # 一坨逻辑
        if p1:
            p1.next = p3
        else:
            dummy.next = p3
        if p2:
            p2.next = p4
        return head

class Solution:
   def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
       pre = None
       cur = head
       i = 0
       p1 = p2 = p3 = p4 = None
       dummy = ListNode(0)
       dummy.next = head
       # 一坨逻辑
       if p1:
           p1.next = p3
       else:
           dummy.next = p3
       if p2:
           p2.next = p4

       return dummy.next

    while cur:
           i += 1
           if i == m - 1:
               p1 = cur
           next = cur.next
           if m < i <= n:
               cur.next = pre

           if i == m:
               p2 = cur
               p2.next = None

           if i == n:
               p3 = cur

           if i == n + 1:
               p4 = cur

           pre = cur
           cur = next

/*
 * @lc app=leetcode id=92 lang=javascript
 *
 * [92] Reverse Linked List II
 *
 * https://leetcode.com/problems/reverse-linked-list-ii/description/
 */
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} m
 * @param {number} n
 * @return {ListNode}
 */
var reverseBetween = function (head, m, n) {
  // 虚拟节点，简化操作
  const dummyHead = {
    next: head,
  };

  let cur = dummyHead.next; // 当前遍历的节点
  let pre = cur; // 因为要反转，因此我们需要记住前一个节点
  let index = 0; // 链表索引，用来判断是否是特殊位置（头尾位置）

  // 上面提到的四个特殊节点
  let p1 = (p2 = p3 = p4 = null);

  while (cur) {
    const next = cur.next;
    index++;

    // 对 (m - n) 范围内的节点进行反转
    if (index > m && index <= n) {
      cur.next = pre;
    }

    // 下面四个if都是边界, 用于更新四个特殊节点的值
    if (index === m - 1) {
      p1 = cur;
    }
    if (index === m) {
      p2 = cur;
    }

    if (index === n) {
      p3 = cur;
    }

    if (index === n + 1) {
      p4 = cur;
    }

    pre = cur;

    cur = next;
  }

  // 两个链表合并起来
  (p1 || dummyHead).next = p3; // 特殊情况需要考虑
  p2.next = p4;

  return dummyHead.next;
};

class Solution:
    def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
        if not head.next or n == 1:
            return head
        dummy = ListNode()
        dummy.next = head
        pre = None
        cur = head
        i = 0
        p1 = p2 = p3 = p4 = None
        while cur:
            i += 1
            next = cur.next
            if m < i <= n:
                cur.next = pre
            if i == m - 1:
                p1 = cur
            if i == m:
                p2 = cur
            if i == n:
                p3 = cur
            if i == n + 1:
                p4 = cur
            pre = cur
            cur = next
        if not p1:
            dummy.next = p3
        else:
            p1.next = p3
        p2.next = p4
        return dummy.next