合并两个有序的单链表，合并之后的链表依然有序

 定义节点 

class ListNode {var next: ListNode = _var x: Int = _def this(x: Int) = {thisthis.x = x}override def toString: String = s"x=>$x"
}

 定义方法 

class LinkedList {var head = new ListNode(0)def getHead(): ListNode = this.headdef add(listNode: ListNode): Unit = {var temp = this.headwhile (temp.next != null) {temp = temp.next}temp.next = listNode}def list(listNode: ListNode): Unit = {var temp = head.nextwhile (temp.next != null) {println(s"$temp")temp = temp.next}}def compair(head1: ListNode, head2: ListNode): ListNode = {var node1 = head1.nextvar node2 = head2.next//定义输出节点val head = new ListNode()var current = new ListNode()head.next = current //将head的next指向currentwhile (node1 != null && node2 != null) {if (node1.x < node2.x) {current.x = node1.xcurrent.next = new ListNode()current = current.nextnode1 = node1.next} else {current.x = node2.xcurrent.next = new ListNode()current = current.nextnode2 = node2.next}}while (node1 != null) {current.x = node1.xcurrent.next = new ListNode()current = current.nextnode1 = node1.next}while (node2 != null) {current.x = node2.xcurrent.next = new ListNode()current = current.nextnode2 = node2.next}head}}

主函数

package Algotithmobject LinkedListDemo {def main(args: Array[String]): Unit = {val list1 = new LinkedListfor (i <- 1 to 5 by 2) {val node = new ListNode(i)list1.add(node)}println(s"list1=========")list1.list(list1.getHead())val list2 = new LinkedListfor (i <- 1 to 10 by 3) {val node = new ListNode(i)list2.add(node)}println(s"list2=========")list2.list(list2.getHead())println(s"=====================")val list3 = new LinkedListval node = list3.compair(list1.getHead(), list2.getHead())list3.head = node // 将 node 赋值给 list3 的头节点list3.list(node)}}

结果

总结

 1、定义方法时，需要重新new节点，将值付给该节点

 2、调用方法得到节点之后，需要将该节点赋值给 list 的头节点