《LeetCode-0002》两数相加-Add Two Numbers

https://leetcode-cn.com/problems/add-two-numbers/

题目

给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。

如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外，这两个数都不会以 0 开头。

示例

1
2
3

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807

思路

题目中加粗关键字注意下：

非空
逆序
一位

解释：
链表从左往右依次是所待变数字的低位到高位， 2 -> 4 -> 3 代表342， 5 -> 6 -> 4 代表465，所以没有必要将链表转成数字再进行相加，也没法那样做，应为如果链表相当之长，会超过整形的范围，所以就从低位往高位挨个算就行了。
针对单次计算x+y=zx+y=zx+y=z，只需要考虑是否向前进位的问题即可。

第一次提交

/**
 * 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 result = new ListNode(0);
        ListNode current1 = l1, current2 = l2, currentResult = result;
        int carry = 0;
        while (current1 != null && current2 != null) {
            int val = (current1.val + current2.val + carry) % 10;
            currentResult.next = new ListNode(val);
            carry = (current1.val + current2.val + carry) / 10;
            current1 = current1.next;
            current2 = current2.next;
            currentResult = currentResult.next;
        }
        while (current1 != null) {
            int val = (current1.val + carry) % 10;
            currentResult.next = new ListNode(val);
            carry = (current1.val + carry) / 10;
            current1 = current1.next;
            currentResult = currentResult.next;
        }
        while (current2 != null) {
            int val = (current2.val + carry) % 10;
            currentResult.next = new ListNode(val);
            carry = (current2.val + carry) / 10;
            current2 = current2.next;
            currentResult = currentResult.next;
        }
        return result.next;
    }
}

第一次提交未通过

测试用例	我的结果	预期结果
［5］,［5］	［0］	［0，1］

忽略了上述这种，结果位数比加数位数多且存在进位的情况

第二次提交

/**
 * 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 result = new ListNode(0);
        ListNode current1 = l1, current2 = l2, currentResult = result;
        int carry = 0;
        while (current1 != null && current2 != null) {
            int val = (current1.val + current2.val + carry) % 10;
            currentResult.next = new ListNode(val);
            carry = (current1.val + current2.val + carry) / 10;
            current1 = current1.next;
            current2 = current2.next;
            currentResult = currentResult.next;
        }
        while (current1 != null) {
            int val = (current1.val + carry) % 10;
            currentResult.next = new ListNode(val);
            carry = (current1.val + carry) / 10;
            current1 = current1.next;
            currentResult = currentResult.next;
        }
        while (current2 != null) {
            int val = (current2.val + carry) % 10;
            currentResult.next = new ListNode(val);
            carry = (current2.val + carry) / 10;
            current2 = current2.next;
            currentResult = currentResult.next;
        }
        if(carry!=0) {
            currentResult.next = new ListNode(carry);
        }
        return result.next;
    }
}