Problem: 2. 两数相加
Code
⏰ 时间复杂度: O ( n ) O(n) O(n)
🌎 空间复杂度: O ( n ) O(n) O(n)
Java
/**
* 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 res = new ListNode();
ListNode cur = res;
int c = 0;//保存进位信息
while (l1 != null && l2 != null)//进行两个数最低位的相加
{
int x = (l1.val + l2.val + c) % 10;
cur.next = new ListNode(x);
cur = cur.next;
c = (l1.val + l2.val + c) / 10;
l1 = l1.next;
l2 = l2.next;
}
while (l1 != null)//处理数位较多的数 和 进位
{
int x = (l1.val + c) % 10;
c = (l1.val + c) / 10;
cur.next = new ListNode(x);
cur = cur.next;
l1 = l1.next;
}
while (l2 != null)//处理数位较多的数 和 进位
{
int x = (l2.val + c) % 10;
c = (l2.val + c) / 10;
cur.next = new ListNode(x);
cur = cur.next;
l2 = l2.next;
}
if (c != 0)//处理最后一个进位
cur.next = new ListNode(c);
return res.next;
}
}