[LeetCode][LintCode] Add Two Numbers
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.
思路:
这道题麻烦的是有carry,需要考虑几种情况:
1) l1 和 l2 一样长,最后一位不需要进位。
2) l1/l2 更长。
3) l1 和 l2 一样长,最后一位还需要进位 。
这道题需要是学习的是,写的要clean一点
/** * 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) { ListNode dummy = new ListNode(0); ListNode head = dummy; int carry = 0; while (l1 != null || l2 != null) { int v1 = l1 == null ? 0 : l1.val; int v2 = l2 == null ? 0 : l2.val; int sum = (v1 + v2 + carry) % 10; carry = (v1 + v2 + carry) / 10; ListNode node = new ListNode(sum); head.next = node; l1 = l1 == null ? null : l1.next; l2 = l2 == null ? null : l2.next; head = head.next; } if (carry != 0) { head.next = new ListNode(carry); } return dummy.next; } }
posted on 2018-02-21 12:50 codingEskimo 阅读(117) 评论(0) 编辑 收藏 举报