程序员代码面试指南第二版 21.两个链表生成相加链表

​​welcome to my blog​​

程序员代码面试指南第二版 21.两个链表生成相加链表

题目描述

假设链表中每一个节点的值都在 0 - 9 之间，那么链表整体就可以代表一个整数。
给定两个这种链表，请生成代表两个整数相加值的结果链表。
例如：链表 1 为 9->3->7，链表 2 为 6->3，最后生成新的结果链表为 1->0->0->0。

输入描述:
第一行两个整数 n 和 m，分别表示两个链表的长度。

第二行 n 个整数 ai 表示第一个链表的节点。

第三行 m 个整数 bi 表示第二个链表的节点。

输出描述:
输出一行整数表示结果链表。

示例1

输入
3 2
9 3 7
6 3

输出
1 0 0 0

第一次做; 跟LeetCode第2题基本一样, 但是比LeetCode第2题难, 因为需要先对链表进行翻转; 需要随时考虑进位; 反转链表的while循环内部有4行代码; 技巧:创建临时节点

import java.util.Scanner;

public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        String str = sc.nextLine();
        int n = Integer.parseInt(str.split(" ")[0]);
        int m = Integer.parseInt(str.split(" ")[1]);
        //input check
        if(n<0 || m<0)
            return;
        String[] linkedList1 = sc.nextLine().split(" ");
        String[] linkedList2 = sc.nextLine().split(" ");
        //生成链表
        ListNode head1 = new ListNode(Integer.parseInt(linkedList1[0]));
        ListNode curr = head1;
        for(int i=1; i<n; i++){
            curr.next = new ListNode(Integer.parseInt(linkedList1[i]));
            curr = curr.next;
        }
        ListNode head2 = new ListNode(Integer.parseInt(linkedList2[0]));
        curr = head2;
        for(int i=1; i<m; i++){
            curr.next = new ListNode(Integer.parseInt(linkedList2[i]));
            curr = curr.next;
        }
        //execute
        //反转链表1
        ListNode left = null, right;
        curr = head1;
        while(curr!=null){
            //save next
            right = curr.next;
            //change
            curr.next = left;
            //update
            left = curr;
            curr = right;
        }
        //保存反转后的头结点
        head1 = left;
        //反转链表2
        left = null;
        curr = head2;
        while(curr!=null){
            //save
            right = curr.next;
            //change
            curr.next = left;
            //update
            left = curr;
            curr = right;
        }
        //保存反转后的头结点
        head2 = left;
        //开始相加
        ListNode p1 = head1, p2 = head2;
        //临时头结点
        ListNode temp = new ListNode(0);
        left = temp;
        int carry = 0, sum;
        while(p1!=null && p2!=null){
            sum = (p1.val + p2.val + carry) % 10;
            carry = (p1.val + p2.val + carry) / 10;
            left.next = new ListNode(sum);
            //update
            left = left.next;
            p1 = p1.next;
            p2 = p2.next;
        }
        while(p1!=null){
            sum = (p1.val + carry) % 10;
            carry = (p1.val + carry) / 10;
            left.next = new ListNode(sum);
            //update
            left = left.next;
            p1 = p1.next;
        }
        while(p2!=null){
            sum = (p2.val + carry) % 10;
            carry = (p2.val + carry) / 10;
            left.next = new ListNode(sum);
            //update
            left = left.next;
            p2 = p2.next;
        }
        if(carry==1){
            left.next  = new ListNode(1);
        }
        //反转结果链表
        left = null;
        curr = temp.next;
        while(curr!=null){
            //save
            right = curr.next;
            //change
            curr.next = left;
            //update
            left = curr;
            curr = right;
        }
        //保存结果链表的头
        temp = left;
        //反转链表1
        right = null;
        curr = head1;
        while(curr!=null){
            //save
            left = curr.next;
            //change
            curr.next = right;
            //update
            right = curr;
            curr = left;
        }
        //保存链表1的头
        head1 = right;
        //反转链表2
        right = null;
        curr = head2;
        while(curr!=null){
            //save
            left = curr.next;
            //change
            curr.next = right;
            //update
            right = curr;
            curr = left;
        }
        //保存链表2的头
        head2 = right;
        //打印结果
        curr = temp;
        while(curr!=null){
            System.out.print(curr.val+" ");
            curr = curr.next;
        }
    }
    public static class ListNode{
        int val;
        ListNode next;
        ListNode(int val){
            this.val = val;
        }
    }
}