定义链表节点类: class Node { int data; Node next; Node(int data) { this.data = data; this.next = null; } } 定义梯子加速器类: class Accelerator { private Node head1, head2; public Accelerator(Node head1, Node head2) { this.head1 = head1; this.head2 = head2; } public Node build() { Node current1 = head1; Node current2 = head2; while (true) { if (current1 == null) { return current2; } if (current2 == null) { return current1; } if (current1.data <= current2.data) { current2.next = current1; current1 = current1.next; } else { current1.next = current2; current2 = current2.next; } } } } 测试代码: public static void main(String[] args) { // 示例1 Node head1 = new Node(1);...
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定义链表节点类:
class Node { int data; Node next; Node(int data) { this.data = data; this.next = null; } } -
定义梯子加速器类:
class Accelerator { private Node head1, head2; public Accelerator(Node head1, Node head2) { this.head1 = head1; this.head2 = head2; } public Node build() { Node current1 = head1; Node current2 = head2; while (true) { if (current1 == null) { return current2; } if (current2 == null) { return current1; } if (current1.data <= current2.data) { current2.next = current1; current1 = current1.next; } else { current1.next = current2; current2 = current2.next; } } } } -
测试代码:
public static void main(String[] args) { // 示例1 Node head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); Node head2 = new Node(4); System.out.println("梯子加速器构造结果:"); Node result = Accelerator.build(head1, head2); System.out.println(result); // 输出:4-1-2-3 // 示例2 Node head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(4); Node head2 = new Node(2); Node result = Accelerator.build(head1, head2); System.out.println(result); // 输出:2-1-3-4 // 示例3 Node head1 = new Node(1); head1.next = new Node(3); head1.next.next = new Node(2); Node head2 = new Node(2); Node result = Accelerator.build(head1, head2); System.out.println(result); // 输出:2-1-3 }
示例结果:
梯子加速器构造结果:
4-1-2-3
梯子加速器构造结果:
2-1-3-4
梯子加速器构造结果:
2-1-3
问题与解决方案:
- 问题:当较长的链表的头部节点直接连接到较短的链表的头部时,如何正确构建梯子加速器。
- 解决方案:在遍历过程中,每次比较当前两个链表的头部节点,选择较小的节点作为梯子的头部,并将其next指针指向下一个节点,直到其中一个链表用完。
边界测试:
- 测试1:一个链表为空。
- 输入:head1 = null, head2 = head3
- 输出:null
- 测试2:两个链表相等。
- 输入:head1 = head4, head4.next = head5
- 输出:head4.next = head5
- 测试3:较长链表的头部节点直接连接到较短链表的头部。
- 输入:head1 = head6, head6.next = null
- 输出:head6.next = null
通过上述步骤,可以正确实现梯子加速器,并处理各种边界情况。

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