Add Two Numbers

今天是一道有关链表的题目，来自LeetCode，难度为Medium，Acceptance为21%。

题目如下

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.
Example
Given 7->1->6 + 5->9->2. That is, 617 + 295.
Return 2->1->9. That is 912.
Given 3->1->5 and 5->9->2, return 8->0->8.

解题思路及代码见阅读原文

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解题思路

首先，该题的难度为中等，但是通过率还是很高的，相信很多同学都能做出来。思路也较为简单，每个节点相加，注意进位。唯一需要注意的是细节，保证无bug。

• 第一个细节是，这里不需要没计算一次就生成一个新节点，使用链表中的节点效率更高。

• 第二个细节是最高位的进位，即当最高位有进位时，要生成一个新节点插入链表结尾。

注意了上述细节后应该就没有太大问题了，下面我们来看代码。

代码如下

Java版

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;      
 *     }
 * }
 */
public class Solution {
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2 
     */
    public ListNode addLists(ListNode l1, ListNode l2) {
        // write your code here
        if(null == l1)
            return l2;
        if(null == l2)
            return l1;
        int carry = 0;
        ListNode head = l1, prev = l1;
        while(l1 != null && l2 != null) {
            prev = l1;
            int val = (l1.val + l2.val + carry) % 10;
            carry = (l1.val + l2.val + carry) / 10;
            l1.val = val;
            l1 = l1.next;
            l2 = l2.next;
        }
        if(l2 != null) {
            prev.next = l2;
            l1 = l2;
        }
        while(l1 != null && carry != 0) {
            prev = l1;
            int val = (l1.val + carry) % 10;
            carry = (l1.val + carry) / 10;
            l1.val = val;
            l1 = l1.next;
        }
        
        if(carry != 0) {
            ListNode tail = new ListNode(carry);
            prev.next = tail;
        }
        return head;
    }
}

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