2. Add two numbers (M)

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本文详细解析了如何使用链表结构表示的两个非负整数进行相加运算，通过逆序存储的链表节点，逐位计算两数之和并处理进位，最终返回结果链表。

Add two numbers (M)

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.

题意

以各位数倒序排列链表的形式给定两个整数，要求计算出两者之和，并同样以各位数倒序链表的形式返回结果。

思路

从两链表的个位数开始，依次计算各位数之和并判断记录进位。最后注意对最高位的进位进行处理。(注意先转换为两个整数再相加的方法会存在测试整数值超过上限的问题)

代码实现 - Java

/**
 * 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 head = new ListNode(0);        // 创建头结点便于输出结果
        ListNode cur = head;
        int carry = 0;
        int a = 0, b = 0, c = 0;
        
        while (l1 != null || l2 != null) {
            a = (l1 == null)? 0 : l1.val;
            b = (l2 == null)? 0 : l2.val;
            c = a + b + carry;              
            carry = c / 10;             // 判断进位
            c = c % 10;
            cur.next = new ListNode(c);
            cur = cur.next;
            l1 = (l1 == null)? null : l1.next;
            l2 = (l2 == null)? null : l2.next;
        }
        // 最高位进位情况处理
        if (carry != 0) {
            cur.next = new ListNode(carry);
        }
        
        return head.next;
    }
}