反转一个单链表。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?
我觉得这道题微妙的地方在于考查递归的思想。目前只用迭代的方法做出了,递归还没想明白,日后补上。
python(迭代):
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if head.next == None:
return head
return
C++(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) {
ListNode* end = NULL;
while(head){
ListNode* Next = head->next;
head->next = end;
end = head;
if(Next) head = Next;
else break;
}
return head;
}
};