题解 | 反转链表
反转链表
https://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca
/**
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param head ListNode类
* @return ListNode类
*/
struct ListNode* ReverseList(struct ListNode* head ) {
// write code here
if (head == NULL || head->next == NULL) {
return head;
}
struct ListNode* p, *q;
p = head->next;
head->next = NULL;//反转的第一步就把原头节点的next置空(因为它最终会是尾节点)
while (p) {
q = p->next;
p->next = head;
head = p;
p = q;
}
return head;
}
#头插法#
