leetcode-92：反转链表 II

92. 反转链表 II

给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。

示例 1：

输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]

示例 2：

输入：head = [5], left = 1, right = 1
输出：[5]

提示：

• 链表中节点数目为 n
• 1 <= n <= 500
• -500 <= Node.val <= 500
• 1 <= left <= right <= n

• /**
   * Definition for singly-linked list.
   * struct ListNode {
   *     int val;
   *     struct ListNode *next;
   * };
   */
  void reverse(struct ListNode *p, struct ListNode *q){
      if (p == q) return;
      struct ListNode *tail = p->next;
      reverse(tail, q);
      p->next = tail->next;
      tail->next = p;
      return;
  }
  
  struct ListNode* reverseBetween(struct ListNode* head, int left, int right){
      struct ListNode new_head, *p = &new_head, *q = &new_head;
      new_head.next = head;
      while (left - 1 > 0) p = p->next, left--;
      while (right > 0) q = q->next, right--;
      reverse(p->next, q);
      p->next = q;
      return new_head.next;
  }

• /**
   * Definition for singly-linked list.
   * struct ListNode {
   *     int val;
   *     struct ListNode *next;
   * };
   */
  
  
  struct ListNode* reverseBetween(struct ListNode* head, int left, int right){
      if (left == 1 && right == 1) return head;
      if (left != 1) {
          head->next = reverseBetween(head->next, left - 1, right - 1);
      }else {
          struct ListNode *new_head, *tail = head->next;
          new_head = reverseBetween(head->next, left, right - 1);
          head->next = tail->next;
          tail->next = head;
          head = new_head;
      }
      return head;
  }