LeetCode题库整理【Java】
2.两数相加
题目:给出两个 非空 的链表用来表示两个非负的整数。其中,它们各自的位数是按照 逆序 的方式存储的,并且它们的每个节点只能存储 一位 数字。
如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。
您可以假设除了数字 0 之外,这两个数都不会以 0 开头。
示例1:
输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出:7 -> 0 -> 8
原因:342 + 465 = 807
示例2:
输入:(9) + (1 -> 9-> 9-> 9-> 9-> 9 -> 9)
输出:1 -> 0 -> 0
原因:9 + 9999991 = 10000000
class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode head1 = l1;ListNode head2 = l2;ListNode head3 = new ListNode(0);ListNode result =head3;//进位标志boolean flag = false;while(head1!=null || head2!=null) {//获取对应位置的值然后相加int x = (head1!=null) ? head1.val : 0;int y = (head2!=null) ? head2.val : 0;int sum = flag ? x+y+1 : x+y;//判断是否有新的进位if( (sum/10) != 0) {sum %= 10;flag = true;}elseflag = false;//新增节点result.next = new ListNode(sum % 10);result = result.next;if (head1 != null)head1 = head1.next;if (head2 != null)head2 = head2.next; }//处理最后一位相加之后的进位if(flag)result.next=new ListNode(1);return head3.next;}
}
在自己的电脑软件上做测试时,可以将ListNode类写为内部类,在public ListNode addTwoNumbers(ListNode l1,ListNode l2) { }方法体中补全代码,在主函数main()入口中输入链表l1和l2的内容并生成链表,并将输出结果l3的内容打印出来。
完整的Java测试代码如下:
/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode() {}* ListNode(int val) { this.val = val; }* ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
package leet.code;import java.util.Scanner;public class AddTwoNumbers {@SuppressWarnings("resource")public static void main(String[] args) {// TODO Auto-generated method stubAddTwoNumbers add = new AddTwoNumbers(); AddTwoNumbers.ListNode head1 = add.new ListNode(0);AddTwoNumbers.ListNode l1=head1;AddTwoNumbers.ListNode head2 = add.new ListNode(0);AddTwoNumbers.ListNode l2=head2;AddTwoNumbers.ListNode head3 = add.new ListNode(0);Scanner sc = new Scanner(System.in);//在控制台输入链表l1各个节点的值,以逗号隔开String str1 = sc.nextLine();String[] num1 = str1.split(",");//用逗号隔开int[] num1int = new int[num1.length];for(int i=0;i<num1int.length;i++) {num1int[i] = Integer.parseInt(num1[i]);l1.next = add.new ListNode(num1int[i]);l1 = l1.next;} head1 = head1.next;//在控制台输入链表l2各个节点的值,以逗号隔开String str2 = sc.nextLine();String[] num2 = str2.split(",");//用逗号隔开int[] num2int = new int[num2.length];for(int i=0;i<num2int.length;i++) {num2int[i] = Integer.parseInt(num2[i]);l2.next = add.new ListNode(num2int[i]);l2 = l2.next;}head2 = head2.next;head3 = add.addTwoNumbers(head1, head2);while(head3!=null) {System.out.println(head3.val); head3=head3.next;}}public class ListNode { //建立内部类ListNodeint val;ListNode next;ListNode() {}ListNode(int val) { this.val = val; }ListNode(int val, ListNode next) { this.val = val; this.next = next; }}public ListNode addTwoNumbers(ListNode l1,ListNode l2) {ListNode head1 = l1;ListNode head2 = l2;ListNode head3 = new ListNode(0);ListNode result =head3;//进位标志boolean flag = false;while(head1!=null || head2!=null) {//获取对应位置的值然后相加int x = (head1!=null) ? head1.val : 0;int y = (head2!=null) ? head2.val : 0;int sum = flag ? x+y+1 : x+y;//判断是否有新的进位if( (sum/10) != 0) {sum %= 10;flag = true;}elseflag = false;//新增节点result.next = new ListNode(sum % 10);result = result.next;if (head1 != null)head1 = head1.next;if (head2 != null)head2 = head2.next; }//处理最后一位相加之后的进位if(flag)result.next=new ListNode(1);return head3.next;}
}
参考网址:https://www.cnblogs.com/mfrank/p/10472639.html
该文章中还给出了很多人可能会踩坑的解题思路,并且给出了代码,也写得很有趣