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LeetCode No2 Add Two Numbers

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题目

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 contains a single digit. Add the two numbers and return the sum as a linked list. 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。这些数字以相反的顺序存储，它们的每个节点都包含一个数字。将这两个数字相加并将总和作为链接列表返回。您可以假设这两个数字不包含任何前导零，除了数字 0 本身。

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public static class ListNode {
    int val;
    ListNode next;
    ListNode() {
    }
    ListNode(int val) {
        this.val = val;
    }
    ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }
}

示例

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Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]
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Input: [2,3,4],[5,6,4]
Output: [7,9,8]
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Input: [2,4,4],[5,6,4]
Output: [7,0,9]

思路

新增两个相同ListNode，一个用于遍历取值，一个用于返回值。

如果val大于10，应该在next结点存为1；

解法1–自己的求解

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public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode list = new ListNode();
    ListNode list2 = list;
    int i;
    if (l1==null && l2==null) return null;
    if (l1 == null) return l2;
    if (l2 == null) return l1;
    for (i = 0; i<= 99;i++) {
        list.val = l1.val + l2.val + list.val;
        if (list.val >= 10) {
            list.val = list.val - 10;
            list.next = new ListNode(1);
        } else {
            list.next = new ListNode();
        }
        if (l1.next == null && l2.next == null) {
            if (list.next.val == 0) {
                list.next = null;
            }
            break;
            } else if (l1.next == null) {
                l1.val = 0;
                l2=l2.next;
            } else if (l2.next == null) {
                l2.val = 0;
                l1=l1.next;
            } else {
                l1=l1.next;
                l2=l2.next;
            }
            list = list.next;
        }
    return list2;
}

缺点：考虑的特殊情况太多，判断性语句过多，导致代码冗余，没有将特殊情况归到一般情况中。

解法2-标准解法

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public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int carry = 0;
        ListNode head = new ListNode(0), p = head;
        while (l1 != null || l2 != null || carry != 0) {
            int sum = carry;
            if (l1 != null) { sum += l1.val; l1 = l1.next; } 
            if (l2 != null) { sum += l2.val; l2 = l2.next; }
            p.next = new ListNode(sum % 10);
            p = p.next;
            carry = sum / 10;
        }
        return head.next;
    }