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#