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LeetCode_2(两数相加)
2022-07-05 13:51:00 【***】
题目描述:
给你两个 非空 的链表,表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的,并且每个节点只能存储 一位 数字。
请你将两个数相加,并以相同形式返回一个表示和的链表。
你可以假设除了数字 0 之外,这两个数都不会以 0 开头。
示例 1:
输入:l1 = [2,4,3], l2 = [5,6,4]
输出:[7,0,8]
解释:342 + 465 = 807.
示例 2:
输入:l1 = [0], l2 = [0]
输出:[0]
示例 3:
输入:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出:[8,9,9,9,0,0,0,1]
提示:
每个链表中的节点数在范围 [1, 100] 内
0 <= Node.val <= 9
题目数据保证列表表示的数字不含前导零
/**
* 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; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode head=null,l=null; //创建头节点和指针节点
int flag=0; //进位标志
while(l1!=null||l2!=null){
//遍历两个单链表
int n1=l1!=null?l1.val:0; //取出第一个链表当前节点的值,若当前节点为空,则补零
int n2=l2!=null?l2.val:0; //同上
int cur=n1+n2+flag; //计算结果链表应当填入的值
if(head==null){
//若为第一个值,则创建节点并连接头指针
l=new ListNode(cur%10);
head=l;
}else{
//创建新节点填入结果并将指针后移
l.next=new ListNode(cur%10);
l=l.next;
}
flag=cur/10; //判断是否需要进位
if(l1!=null)l1=l1.next; //指针后移前判断一下,以免指针异常
if(l2!=null)l2=l2.next;
}
if(flag>0)l.next=new ListNode(flag);
return head;
}
}
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