面试手撕代码(2)-链表的反转

listnode reverse(listnode head)
{
if (head == NULL)
return head;
listnode p, q, pre;
p = head->next;
q = NULL;
head->next == NULL;
while (p)
{
pre = p->next;
p->next = q;
q = p;
p = pre;
}
head->next = q;
return head;
}

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转载自blog.csdn.net/m0_37947204/article/details/79960200