LeetCode 2 -- Add Two Numbers

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

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在码的时候需要注意各个边界问题，考虑到零输入，单输入，单进位的可能性，反复修改，例如：[0][1]，[5][5]，[1,8][0]

在编程之前要先想好算法以及过程，先谋而后动。

可以用三指针同时进行，代码会简洁很多，不要破坏原有的链表，这是一个很好的习惯。

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
       int carry =0;
 
        ListNode newHead = new ListNode(0);
        ListNode p1 = l1, p2 = l2, p3=newHead;
 
        while(p1 != null || p2 != null){
            if(p1 != null){
                carry += p1.val;
                p1 = p1.next;
            }
 
            if(p2 != null){
                carry += p2.val;
                p2 = p2.next;
            }
 
            p3.next = new ListNode(carry%10);
            p3 = p3.next;
            carry /= 10;
        }
 
        if(carry==1) 
            p3.next=new ListNode(1);
 
        return newHead.next;
    }
}

 下面的代码的思路是将为空的链表的值设为0：

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        
        ListNode result = new ListNode(0);
        ListNode start = result;
        int temp = 0;
        ListNode t1 = l1;
        ListNode t2 = l2;
        int t1v;
        int t2v;
        for (; t1 != null || t2 != null;) {
            result.next = new ListNode(0);
            result = result.next;
            if (t1 == null)
                t1v = 0;
            else {
                t1v = t1.val;
                t1 = t1.next;
            }
            if (t2 == null)
                t2v = 0;
            else {
                t2v = t2.val;
                t2 = t2.next;
            }
            result.val = (t1v + t2v + temp) % 10;
            temp = (t1v + t2v + temp) / 10;

        }
        if (temp != 0) {
            result.next = new ListNode(0);
            result.next.val = temp;
        }
        return start.next;

    }
}