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。

新增一个链表,然后分别采用2个指针指向这两个链表,每次比较链表的值,将较小的那一个存入新链表中。需要主要的是对空链表进行处理,主要还是考察代码的鲁棒性。总共2个链表,数组分别是1,3,5,7,和2, 4, 6,8

 

采用java进行实现。

package com.test.algorithm;




class ListNode {
    int val;
    ListNode next = null;

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

public class Merge {
    
    public static ListNode Merge(ListNode list1,ListNode list2) {
        
        if(list1 == null && list2 == null)
            return null;
        if(list1 == null)
            return list2;
        if(list2 == null)
            return list1;
        
        ListNode newNode = new ListNode(0);
        ListNode tmp = new ListNode(0);
        tmp = newNode;
        if(list1.val < list2.val){
            newNode.val = list1.val;
            list1 = list1.next;
        }
        else{
            newNode.val = list2.val;
            list2 = list2.next;
        }
        //newNode.next = null;
        while(list1!=null && list2!=null){
            if(list1.val < list2.val){
                newNode.next = list1;
                newNode = newNode.next;
                newNode.val = list1.val;
                list1 = list1.next;
            }else{
                newNode.next = list2;
                newNode = newNode.next; 
                newNode.val = list2.val;    
                list2 = list2.next;
            }
        }
        if(list1!=null){
            newNode.next = list1;
            newNode = newNode.next;
            //list1 = list1.next;
        }
        if(list2!=null){
            newNode.next = list2;
            newNode = newNode.next;
            //list2 = list2.next;
        }
        
        return tmp;
    }

    public static void main(String[] args) {
        
        ListNode list1 = new ListNode(1);
        ListNode list2 = new ListNode(2);
        list1.next = null;
        list2.next = null;
        ListNode head1 = list1;
        ListNode head2 = list2;
        
        for(int i = 2 ; i <=8 ; i ++){
            if(i%2!=0)
            {
                ListNode temp = new ListNode(i);
                temp.next = list1.next;
                list1.next = temp;
                list1 = temp;
            }
        }
        for(int i = 4 ; i <=8 ; i ++){
            if(i%2==0)
            {
                ListNode temp = new ListNode(i);
                temp.next = list2.next;
                list2.next = temp;
                list2 = temp;
            }
        }
        
        ListNode newNode = Merge(head1,head2);
        for(int i = 1 ; i <= 8 ; i++){
            System.err.println(newNode.val);
            newNode = newNode.next;
        }
    }
}

 

posted @ 2017-08-03 10:31  Cloud_strife  阅读(1313)  评论(0编辑  收藏  举报