反转一个单链表。
示例:
输入: 1->2->3->4->5->NULL 输出: 5->4->3->2->1->NULL
进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题?
递归:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
if(head==null||head.next==null)
return head;
ListNode h = reverseList(head.next);
head.next.next = head;
head.next = null;
return h;
}
}
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if(head==NULL||head->next==NULL)
return head;
ListNode *h = reverseList(head->next);
head->next->next = head;
head->next = NULL;
return h;
}
};
迭代:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if(head == NULL || head->next == NULL)
return head;
ListNode *p = head;
ListNode *q = head->next;
ListNode *r = head->next->next;
p->next = NULL;
while(r != NULL){
q->next = p;
p = q;
q = r;
r = r->next;
}
q->next = p;
return q;
}
};