标签：git tps lis ddt question nbsp div you single
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
给定两个表示两个非负整数的非空链表。这些数字以反向顺序存储，每个节点都包含一个数字。将这两个数相加，并以链表的形式返回它们的和。
假设这两个数字不包含任何前导零，除了数字0本身。
Example 1:
Input: l1 = [2,4,3], l2 = [5,6,4] Output: [7,0,8] Explanation: 342 + 465 = 807.
Example 2:
Input: l1 = [0], l2 = [0] Output: [0]
Example 3:
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9] Output: [8,9,9,9,0,0,0,1]
Constraints:
[1, 100].0 <= Node.val <= 9
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; } } public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode list = new ListNode(-1); ListNode current = list; int carry = 0; while (l1 != null || l2 != null) { int d1 = l1 == null ? 0 : l1.val; int d2 = l2 == null ? 0 : l2.val; int sum = d1 + d2 + carry; carry = sum >= 10 ? 1 : 0; current.next = new ListNode(sum % 10); current = current.next; if (l1 != null) l1 = l1.next; if (l2 != null) l2 = l2.next; } if (carry == 1) current.next = new ListNode(1); return list.next; }
时间复杂度为O(n)，空间复杂度为O(n)。
标签：git tps lis ddt question nbsp div you single
原文地址：https://www.cnblogs.com/IsMyLucas/p/14025070.html
