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) = {
    this
    this.x = x
  }

  override def toString: String = s"x=>$x"
}

 定义方法 

class LinkedList {
  var head = new ListNode(0)

  def getHead(): ListNode = this.head

  def add(listNode: ListNode): Unit = {
    var temp = this.head
    while (temp.next != null) {
      temp = temp.next
    }
    temp.next = listNode
  }

  def list(listNode: ListNode): Unit = {
    var temp = head.next
    while (temp.next != null) {
      println(s"$temp")
      temp = temp.next
    }
  }


  def compair(head1: ListNode, head2: ListNode): ListNode = {
    var node1 = head1.next
    var 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current
    while (node1 != null && node2 != null) {
      if (node1.x < node2.x) {
        current.x = node1.x
        current.next = new ListNode()
        current = current.next
        node1 = node1.next
      } else {
        current.x = node2.x
        current.next = new ListNode()
        current = current.next
        node2 = node2.next
      }
    }
    while (node1 != null) {
      current.x = node1.x
      current.next = new ListNode()
      current = current.next
      node1 = node1.next
    }
    while (node2 != null) {
      current.x = node2.x
      current.next = new ListNode()
      current = current.next
      node2 = node2.next
    }
    head
  }

}

主函数

package Algotithm

object LinkedListDemo {
  def main(args: Array[String]): Unit = {
    val list1 = new LinkedList
    for (i <- 1 to 5 by 2) {
      val node = new ListNode(i)
      list1.add(node)
    }
    println(s"list1=========")
    list1.list(list1.getHead())

    val list2 = new LinkedList
    for (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 LinkedList
    val 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