Leetcode刷题日记 -- Add Two Numbers

贴题：

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

算法思路：

因为是倒序排列的list，所以比较简单，两个链表从头到尾，同一位置的相加，得到的值创建一个新的node即可，使用pre存上一个node，从而将所有node链接起来。

注意事项：

list不一样长会有一个list多出几位，要把这些也要连在上一步创建的新list里，并且要考虑到进位的存在如 9->9->9, 1->0。这样的链表相加最后要多出一个node存放进位

0->0->0->1。

贴代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = null;
        int carry = 0;
        int start = 0;
        int val = 0;
        if(l1 == null)return l2;
        if(l2 == null)return l1;
        ListNode pre = null;
        while(l1 != null && l2 != null){
            val = (l1.val + l2.val + carry)%10;
            carry = (l1.val + l2.val + carry)/10;
            ListNode tmp = new ListNode(val);
            if(start == 0){
                start = 1;
                head = tmp;
                pre = tmp;
            }else{
                pre.next = tmp;
                pre = tmp;
            }
            l1 = l1.next;
            l2 = l2.next;
        }
        while(l1 != null){
            val = (l1.val + carry)%10;
            carry = (l1.val + carry)/10;
            ListNode tmp = new ListNode(val);
            pre.next = tmp;
            pre = tmp;
            l1 = l1.next;
        }
        while(l2 != null){
            val = (l2.val + carry)%10;
            carry = (l2.val + carry)/10;
            ListNode tmp = new ListNode(val);
            pre.next = tmp;
            pre = tmp;
            l2 = l2.next;
        }
        if(carry != 0){
            ListNode tmp = new ListNode(carry);
            pre.next = tmp;
            pre = tmp;
        }
        return head;
    }
}