剑指Offer_编程题_合并两个排序的链表

Posted 2021-02-22 _Phoenix

剑指Offer_编程题_合并两个排序的链表

 

题目描述

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。

 

题目答案，思路

https://www.nowcoder.com/questionTerminal/d8b6b4358f774294a89de2a6ac4d9337?f=discussion

递归版本

链接：https://www.nowcoder.com/questionTerminal/d8b6b4358f774294a89de2a6ac4d9337?f=discussion
来源：牛客网

public ListNode Merge(ListNode list1,ListNode list2) {
       if(list1 == null){
           return list2;
       }
       if(list2 == null){
           return list1;
       }
       if(list1.val <= list2.val){
           list1.next = Merge(list1.next, list2);
           return list1;
       }else{
           list2.next = Merge(list1, list2.next);
           return list2;
       }       
   }

非递归版本

public class Solution {
    public ListNode Merge(ListNode list1,ListNode list2) {
    ListNode  root = new ListNode(-1);
        //临时指向的节点,可以理解为临时的 root 尾节点
        ListNode last = root;
        while (list1 != null && list2 != null) {
            if (list1.val <= list2.val) {
                //重新
                last.next=list1;
                last = list1;
                list1 = list1.next;
            } else {
                last.next = list2;
                last = list2;
                list2 = list2.next;
            }
        }
        //把未结束的链表连接到合并后的链表尾部
        if (list1 != null) {
            last.next = list1;
        }
        if (list2 != null) {
            last.next = list2;
        }
        return root.next;
    }
}

 

 

测试代码

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

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


public class TestListNode {

    //        链接：https://www.nowcoder.com/questionTerminal/d8b6b4358f774294a89de2a6ac4d9337?f=discussion
//        来源：牛客网
//尾插法创建单链表  队列形式先进先出
    public void back(ListNode node, int data) {
        if (data < 10) {
            ListNode next = new ListNode(data);
            next.val = data;
            next.next = null;
            node.next = next;
            back(next,data+3);
        }
    }

    public void back2(ListNode node, int data) {
        if (data < 10) {
            ListNode next = new ListNode(data);
            next.next = null;
            node.next = next;
            back2(next, data=data+2);
        }
    }



    public static void main(String[] args) {
         ListNode headNode;
         ListNode headNode2;
        TestListNode list1 = new TestListNode();
        headNode = new ListNode(3);//头指针
        list1.back(headNode, 4);//后插法

        System.out.println("创建后的链表是：");
//        while (headNode.next != null) {
//            headNode = headNode.next;
//            System.out.print(headNode.val + " ");
//        }
        System.out.println("-----------------------------");
        TestListNode list2 = new TestListNode();
        headNode2 = new ListNode(4);//头指针
        list2.back2(headNode2, 5);//后插法


        System.out.println("创建后的链表是：");
//        while (headNode2.next != null) {
//            headNode2 = headNode2.next;
//            System.out.print(headNode2.val + " ");
//        }

        ListNode merged=  new Solution().Merge(headNode,headNode2);

        System.out.println("merge ：");
        while (merged.next != null) {
            merged = merged.next;
            System.out.print(merged.val + " ");
        }

    }

}