（每日一练c语言）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 n1 <= n <= 500-500 <= Node.val <= 5001 <= left <= right <= n

以下程序实现了这一功能：

#include 
#include 
struct ListNode
{
	int val;
	struct ListNode *next;
};
static struct ListNode *reverseBetween(struct ListNode *head, int m, int n)
{
	int i;
	struct ListNode dummy;
	struct ListNode *prev = &dummy;
	prev->next = head;
	for (i = 1; i < m; i++)
	{
		prev = prev->next;
	}
	struct ListNode *p = prev->next;
	for (i = m; i < n; i++)
	{
		struct ListNode *q = p->next;
		p->next = q->next;
        q->next = prev->next;
        prev->next = q;
	}
	return dummy.next;
}
int main(int argc, char **argv)
{
	if (argc < 3)
	{
		fprintf(stderr, "Usage: ./test m n 1 2 3...n");
		exit(-1);
	}
	int i, count = argc - 3;
	struct ListNode dummy;
	struct ListNode *prev = &dummy;
	struct ListNode *p;
	for (i = 0; i < count; i++)
	{
		p = malloc(sizeof(*p));
		p->val = atoi(argv[i + 3]);
		p->next = NULL;
		prev->next = p;
		prev = p;
	}
	int m = atoi(argv[1]);
	int n = atoi(argv[2]);
	struct ListNode *head = reverseBetween(dummy.next, m, n);
	for (p = head; p != NULL; p = p->next)
	{
		printf("%d ", p->val);
	}
	printf("n");
	return 0;
}