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？
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辅助数组法：首先遍历链表中的元素，并使用额外的数组保存访问的节点值，然后将数组反转，最后使用反转后的数组构建新的链表并返回链表头。
时间复杂度 O ( N ) O(N) O(N)
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
if not head: return None
r = []
cur = head
while cur:
r.append(cur.val)
cur = cur.next
r.reverse()
new = ListNode(-1)
cur = new
for i in r:
cur.next = ListNode(i)
cur = cur.next
return new.next
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直接反转：设置
new作为反转链表的头结点，然后依次访问给定链表中的元素，将其不断的插入到new前，并不断的更新new，最后返回表头new即可时间复杂度 O ( N ) O(N) O(N)
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
if not head: return None
new = None
cur = head
while cur:
node = ListNode(cur.val)
node.next = new
new = node
cur = cur.next
return new
/**
* 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){
return null;
}
ListNode node = null;
ListNode cur = head;
//ListNode temp = null;
while(cur != null){
ListNode temp = new ListNode(cur.val);
temp.next = node;
node = temp;
cur = cur.next;
}
return node;
}
}
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
if not head or not head.next:
return head
nextNode = self.reverseList(head.next)
head.next.next = head
head.next = None
return nextNode
