题解 | #链表相加(二)#
链表相加(二)
https://www.nowcoder.com/practice/c56f6c70fb3f4849bc56e33ff2a50b6b
import java.util.*;
public class Solution {
//反转链表
public ListNode ReverseList(ListNode pHead) {
if (pHead == null)
return null;
ListNode cur = pHead;
ListNode pre = null;
while (cur != null) {
//断开链表,要记录后续一个
ListNode temp = cur.next;
//当前的next指向前一个
cur.next = pre;
//前一个更新为当前
pre = cur;
//当前更新为刚刚记录的后一个
cur = temp;
}
return pre;
}
public ListNode addInList (ListNode head1, ListNode head2) {
//任意一个链表为空,返回另一个
if (head1 == null)
return head2;
if (head2 == null)
return head1;
//反转两个链表
head1 = ReverseList(head1);
head2 = ReverseList(head2);
//添加表头
ListNode res = new ListNode(-1);
ListNode head = res;
//进位符号
int carry = 0;
//只要某个链表还有或者进位还有
while (head1 != null || head2 != null || carry != 0) {
//链表不为空则取其值
int val1 = head1 == null ? 0 : head1.val;
int val2 = head2 == null ? 0 : head2.val;
//相加
int temp = val1 + val2 + carry;
//获取进位
carry = temp / 10;
temp %= 10;
//添加元素
head.next = new ListNode(temp);
head = head.next;
//移动下一个
if (head1 != null)
head1 = head1.next;
if (head2 != null)
head2 = head2.next;
}
//结果反转回来
return ReverseList(res.next);
}
}
