合并两个排序的链表

题目 :输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

思想 :类似于排有序数组,,,定义新node/数组.每次遍历将小的放在node/数组中,移动小的node/数组的游标

代码

package com.algorithm.offer;

import com.sun.xml.internal.bind.v2.model.core.ID;

public class Merge {
   static class  ListNode {
        int val;
        ListNode next = null;

        ListNode(int val) {
            this.val = val;
        }
    }

    //输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。
    public  static ListNode Merge(ListNode list1,ListNode list2) {
        if (list1==null){
            return list2;
        }
        if (list2==null){
            return list1;
        }

        ListNode mergeHead =null;
        ListNode curr=null;

        while (list1!=null&&list2!=null){
            if (list1.val<=list2.val){
                if(mergeHead == null){
                    mergeHead = curr = list1;
                }else{
                    curr.next = list1;
                    curr = curr.next;
                }
                list1 = list1.next;
            }else {
                if(mergeHead == null){
                    mergeHead = curr = list2;
                }else{
                    curr.next = list2;
                    curr = curr.next;
                }
                list2 = list2.next;
            }
        }
        if (list1!=null){
            curr.next=list1;
        }else {
            curr.next=list2;
        }
        return mergeHead;
    }

    public static void main(String[] args){
        ListNode l1=new ListNode(1);
        ListNode l2=new ListNode(3);
        ListNode l3=new ListNode(4);
        ListNode l6=new ListNode(5);
        l1.next=l2;
        l2.next=l3;
        l3.next=l6;

        ListNode l4=new ListNode(2);
        ListNode l5=new ListNode(3);
        ListNode l8=new ListNode(3);
        l4.next=l5;
        l5.next=l8;
        ListNode node = Merge(l1, l4);
        while (node!=null){
            System.out.print(node.val+" ");
            node=node.next;
        }
   }
}