给您一个不可变的链表,使用下列接口逆序打印每个节点的值:
ImmutableListNode
: 描述不可变链表的接口,链表的头节点已给出。
您需要使用以下函数来访问此链表(您 不能 直接访问 ImmutableListNode
):
ImmutableListNode.printValue()
:打印当前节点的值。ImmutableListNode.getNext()
:返回下一个节点。
输入只用来内部初始化链表。您不可以通过修改链表解决问题。也就是说,您只能通过上述 API 来操作链表。
示例 1:
输入:head = [1,2,3,4] 输出:[4,3,2,1]
示例 2:
输入:head = [0,-4,-1,3,-5] 输出:[-5,3,-1,-4,0]
示例 3:
输入:head = [-2,0,6,4,4,-6] 输出:[-6,4,4,6,0,-2]
提示:
- 链表的长度在
[1, 1000]
之间。 - 每个节点的值在
[-1000, 1000]
之间。
进阶:
您是否可以:
- 使用常数级空间复杂度解决问题?
- 使用线性级时间复杂度和低于线性级空间复杂度解决问题?
# """
# This is the ImmutableListNode's API interface.
# You should not implement it, or speculate about its implementation.
# """
# class ImmutableListNode:
# def printValue(self) -> None: # print the value of this node.
# def getNext(self) -> 'ImmutableListNode': # return the next node.
class Solution:
def printLinkedListInReverse(self, head: 'ImmutableListNode') -> None:
if head:
self.printLinkedListInReverse(head.getNext())
head.printValue()
/**
* // This is the ImmutableListNode's API interface.
* // You should not implement it, or speculate about its implementation.
* interface ImmutableListNode {
* public void printValue(); // print the value of this node.
* public ImmutableListNode getNext(); // return the next node.
* };
*/
class Solution {
public void printLinkedListInReverse(ImmutableListNode head) {
if (head != null) {
printLinkedListInReverse(head.getNext());
head.printValue();
}
}
}
/**
* // This is the ImmutableListNode's API interface.
* // You should not implement it, or speculate about its implementation.
* class ImmutableListNode {
* public:
* void printValue(); // print the value of the node.
* ImmutableListNode* getNext(); // return the next node.
* };
*/
class Solution {
public:
void printLinkedListInReverse(ImmutableListNode* head) {
if (head) {
printLinkedListInReverse(head->getNext());
head->printValue();
}
}
};
/* Below is the interface for ImmutableListNode, which is already defined for you.
*
* type ImmutableListNode struct {
*
* }
*
* func (this *ImmutableListNode) getNext() ImmutableListNode {
* // return the next node.
* }
*
* func (this *ImmutableListNode) printValue() {
* // print the value of this node.
* }
*/
func printLinkedListInReverse(head ImmutableListNode) {
if head != nil {
printLinkedListInReverse(head.getNext())
head.printValue()
}
}