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 ) {
struct ListNode *current = head;
struct ListNode *prev = NULL;
struct ListNode *next = NULL;
while (current != NULL) {
next = current->next; //保留后面的链表
current->next = prev; //把当前链表指针域的值变成最后一个值，指针域置为空
prev = current; //把第一个值转变成最后一个值赋给prev,
current = next; //当前指针向后移动
}
return prev;
}
查看17道真题和解析