[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

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