题解 | #反转链表#

单链表的翻转 1、head节点存储数据 2、head节点不存储数据 struct ListNode * reverList(struct ListNode *pHead) { struct ListNode *curr = NULL; struct ListNode *pre = NULL; struct ListNode *next = NULL; temp = pHead; while(temp != NULL) { next = temp->next; //保存好下一个节点 temp->next = pre; //该变当前节点的指向 //向后移项 pre = temp; temp = next; } return pre; }