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
//思路二,原链表翻转
if(head == NULL || head->next == NULL){
return head;
}
//prev 前指针, curt 当前指针, 需要做的事情,将curt 指针指向prev ,prev往后走, 然后curt 往后走,
struct ListNode * prev = NULL;
struct ListNode * curt = head;
while(curt != NULL){
struct ListNode * curtNextTemp = curt->next;
curt->next = prev;
prev = curt;
curt = curtNextTemp;
}
return prev;
}
#链表翻转#
