题解 | #反转链表#
反转链表
https://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca?tpId=308&tqId=23286&ru=/exam/company&qru=/ta/algorithm-start/question-ranking&sourceUrl=%2Fexam%2Fcompany
/** * 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 //使用头插法 struct ListNode* new_Head = (struct ListNode*)malloc(sizeof(struct ListNode)); struct ListNode* p; struct ListNode* r; struct ListNode* s; p = head; new_Head->next = NULL; r = s = new_Head; while(p != NULL){ s = (struct ListNode*)malloc(sizeof(struct ListNode)); s->next = r->next; s->val = p->val; r->next = s; s = r; p = p->next; } return new_Head->next; }