合并两个排序的链表

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题目:输入两个递增排序的链表，合并这两个链表并使新链表中的节点任然是按照递增排序的。

代码如下:

     public class ListDemo {

              public static void main(String[] args) {
       /**
       *
       * 构造链表一
       * */
       ListNode head1 = new ListNode(1);
       head1.next = new ListNode(3);
       head1.next.next = new ListNode(5);
       head1.next.next.next = new ListNode(7);
       head1.next.next.next.next = new ListNode(9);
       head1.next.next.next.next.next = new ListNode(11);
       head1.next.next.next.next.next.next = null;
       /**
       *
       * 构造链表二
       * */
       ListNode head2 = new ListNode(2);
       head2.next = new ListNode(4);
       head2.next.next = new ListNode(6);
       head2.next.next.next = new ListNode(7);
       head2.next.next.next.next = new ListNode(8);
       head2.next.next.next.next.next = null;

       printList(head1);
       System.out.println();
       printList(head2);
       ListNode head3 = merge(head1, head2);
       System.out.println();
       printList(head3);
   }

   private static ListNode merge(ListNode head1, ListNode head2) {
       /**
       * 当head1为空时，直接返回head2链接到上一个节点就好
       * */
       if (head1 == null) {
           return head2;
       }
       /**
       * 当head2为空时，直接返回head1链接到上一个节点就好
       * */
       if (head2 == null) {
           return head1;
       }
       ListNode head = null;
       if (head1.value <= head2.value) {
           // 当head1.value较小时，说明head1将为头节点返回
           head = head1;
           head.next = merge(head1.next, head2);
       } else {
           // 当head1.value较小时，说明head1将为头节点返回
           head = head2;
           head.next = merge(head1, head2.next);
       }
       return head;
   }

   private static void printList(ListNode head1) {

       while (head1 != null) {
           System.out.print(head1.value + " ");
           head1 = head1.next;
       }
   }

}

/**
*
* 链表的节点类
* */
class ListNode {
   public int value;
   public ListNode next;

   public ListNode(int value) {
       super();
       this.value = value;
   }

   public ListNode() {
       super();
   }
}

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