2. Add Two Numbers （两数求和）-蒲公英云

2. Add Two Numbers （两数求和）

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

题目大意：两个非负数分别逆序存放在两个单链表中，将两个单链表代表的数求和，并将和以单链表的形式返回。

解题思路：双指针，先遍历完较短的链表，将两个链表对应节点值求和之后的值更新到较长的链表中，较短的链表遍历完后再接着遍历较长链表剩下的部分，此时仅以较长链表中节点的值作为求和加数，但是要注意处理有进位的情况。

源代码：（57 ms，beats 47.31%）

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int lenth1 = 0, lenth2 = 0;
        ListNode p = l1, q = l2, result;
        int curRes;
        // 进位
        int carry = 0;
        while (p != null) {
            lenth1++;
            p = p.next;
        }
        while (q != null) {
            lenth2++;
            q = q.next;
        }
        if (lenth1 < lenth2) {
            result = l2;
            p = l1;
            q = l2;
        } else {
            result = l1;
            p = l2;
            q = l1;
        }
        ListNode preq = null;
        while (p != null) {
            curRes = p.val + q.val + carry;
            if (curRes >= 10) {
                carry = 1;
                curRes = curRes % 10;
            } else {
                carry = 0;
            }
            q.val = curRes;
            preq = q;
            p = p.next;
            q = q.next;
        }
        while (q != null) {
            curRes = q.val + carry;
            if (curRes >= 10) {
                carry = 1;
                curRes = curRes % 10;
            } else {
                carry = 0;
            }
            q.val = curRes;
            preq = q;
            q = q.next;
        }
        if (carry == 1) {
            preq.next = new ListNode(1);
        }
        return result;
    }
}