Reverse a singly linked list.
Example:
Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL
Follow up:
A linked list can be reversed either iteratively or recursively. Could you implement both?
反转链表。有两种方法可以解决:
1.迭代
1 struct ListNode* reverseList(struct ListNode* head) { 2 if(!head) return NULL; 3 struct ListNode *prev=NULL,*next; 4 while(head){ 5 next=head->next; 6 head->next=prev; 7 prev=head; 8 head=next; 9 } 10 return prev; 11 }
2.递归
1 //递归方式 2 Node * reverseList(List head) 3 { 4 //如果链表为空或者链表中只有一个元素 5 if(head == NULL || head->next == NULL) 6 { 7 return head; 8 } 9 else 10 { 11 //先反转后面的链表，走到链表的末端结点 12 Node *newhead = reverseList(head->next); 13 //再将当前节点设置为后面节点的后续节点 14 head->next->next = head; 15 //原本的head指向关系给设置成NULL，不然head会指向后面结点报错 16 head->next = NULL; 17 18 return newhead; 19 } 20 }