You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
翻译:
给定两个非空链表来代表两个非负数,位数按照逆序方式存储,它们的每个节点只存储单个数字。将这两数相加会返回一个新的链表。
你可以假设除了数字 0 之外,这两个数字都不会以零开头。
示例:
输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出:7 -> 0 -> 8
原因:342 + 465 = 807
主要是大数相加的思想,注意判断进位
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null || l2 == null) {
return l1 == null ? l2 : l1;
}
ListNode p = new ListNode(0);
ListNode res = p;//定义结果链表
int up = 0;//进位
int value = 0;//个位数
while(l1 != null && l2 != null){
value = (l1.val + l2.val + up) % 10;
up = (l1.val + l2.val + up) / 10;
p.next = new ListNode(value);
l1 = l1.next;
l2 = l2.next;
p = p.next;
if(l1 == null && l2 == null && up == 0){ //测试用例有一个是{5}{5}结果为[0,1],当l1,l2都加完时,需要判断进位是否为0,如果不是,需要在结果链表里加上新的节点
break;
}
if(l1 == null){
l1 = new ListNode(0);//如果l1已经到最后一位,而l2还没有,那么在l1后面补零,即在数的高位补0
}
if(l2 == null){
l2 = new ListNode(0);
}
if(up != 0){
p.next = new ListNode(up);//如果最高位有进位
}
}
return res.next;
}
}