LeetCode 2. Add Two Numbers [Medium]

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链表加法。

题目来源：https://leetcode.com/problems/add-two-numbers

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

题目难度：Medium

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.

解答1[Java]：

核心思想

代码

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class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = null, tail = null;
        int carry = 0;
        while (l1 != null || l2 != null) {
            int n1 = l1 != null ? l1.val : 0;
            int n2 = l2 != null ? l2.val : 0;
            int sum = n1 + n2 + carry;
            if (head == null) {
                head = tail = new ListNode(sum % 10);
            } else {
                tail.next = new ListNode(sum % 10);
                tail = tail.next;
            }
            carry = sum / 10;
            if (l1 != null) {
                l1 = l1.next;
            }
            if (l2 != null) {
                l2 = l2.next;
            }
        }
        if (carry > 0) {
            tail.next = new ListNode(carry);
        }
        return head;
    }
}

解答2[Java]：递归思路

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public class Problem2Solution2 {
    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) {
        if (l1 == null && l2 == null) {
            return null;
        }

        if (l1 == null) {
            l1 = new ListNode(0);
        }

        if (l2 == null) {
            l2 = new ListNode(0);
        }

        int rval = l1.val + l2.val;
        if (rval >= 10) {
            rval = rval % 10;
            if (l1.next != null) {
                l1.next.val += 1;
            } else {
                l1.next = new ListNode(1);
            }
        }

        return new ListNode(rval, addTwoNumbers(l1.next, l2.next));
    }
}

进阶：链表中的数是正向存储的

思路1：给较短链表的长度和较长链表的长度补齐，补的高位节点的值都设为0。