反转从位置 m 到 n 的链表【进阶】

之前写的代码我实在看不下去了,因为刚才看的时候已经有些地方一下子看不懂了,

总的来说就是太多的特殊处理,虽然不得不这样做,试着给单链表加了个头节点,

代码简洁多了,强烈建议链表的删除,插入和反转都加上个头节点!!

struct ListNode* reverseBetween(struct ListNode* head, int m, int n) {
  if (!head || !head->next || n == m) return head;//预处理
    struct ListNode *L = (struct ListNode *)malloc(sizeof(struct ListNode));
    L->next = head;     //删除节点常用手段考虑加个头指针
    struct ListNode* pre = NULL;
	struct ListNode* p = L;
	int temp_m = m;
	int temp_n = n;
	//将p转移到反转链表的第一个节点,pre保存前一个节点
	while (temp_m-- >= 1) {
		pre = p;
		p = p->next;
	}

	struct ListNode* reverse_head = NULL;
	struct ListNode* reverse_tail = p;
	struct ListNode* reverse_tail_after = NULL;
	struct ListNode* q = NULL;
	//进行反转
	while (temp_n - m > 0) {
		q = p->next;
		reverse_tail_after = q->next;
		p->next = reverse_head;
		reverse_head = p;
		p = q;
		temp_n--;
	}
    
	p->next = reverse_head;
	reverse_head = p;
	pre->next = reverse_head;
    
	reverse_tail->next = reverse_tail_after;
	return L->next;
}

欢迎大家评论!!!

具体不同之处请参见我之前写的博客!!!

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转载自blog.csdn.net/liuxiang15/article/details/82192528