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

感想： 刚开始看到这道题，觉得很简单，跟归并merge的过程比较像，算法复杂度O（n）。可以半小时之内完全可以搞定。可是随着一次次提交出问题，发现似乎没有我想的那么简单...问题其实就出现在各种corner cases没有处理好，which is also 剑指offer多次提到的写代码要注意健壮性....深刻体会到了。

思路： 处理好进位，要自己手动写各种情况的测试用例，确保程序的健壮性...

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    private int getSum(int x, int y, int digit){
        int value = x+y+digit;
        if(value>=10)
            return value-10;
        else
            return value;
    }
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode p = l1;
        ListNode q = l2;
        // new head node for the resultlist
        ListNode newHead = new ListNode(getSum(p.val,q.val,0));
        ListNode pHead = newHead;

        int digit;//处理进位
        if(p.val+q.val-10 >= 0){
            digit = 1;
        }
        else 
            digit = 0;

        p = p.next;
        q = q.next;

        while(p!=null || q != null){
            // 两个链表长度相等时候的处理
            if(p!=null &&q!=null){
                ListNode temp = new ListNode(getSum(p.val, q.val,digit));        
                pHead.next  = temp;
                pHead = pHead.next;
                if(p.val+q.val+digit-10 >= 0){
                    digit = 1;
                }
                else 
                    digit = 0;
                p = p.next;
                q = q.next;
            }
            // 两个链表长度不等时候的处理
            else if(p==null){
                ListNode temp = new ListNode(getSum(0, q.val,digit));    
                pHead.next  = temp;
                pHead = pHead.next;
                if(0+q.val-10+digit >= 0){
                    digit = 1;
                } 
                else{
                    digit = 0;
                }
                q = q.next;
            }
            // 两个链表长度不等时候的处理
            else if(q==null){
                ListNode temp = new ListNode(getSum(p.val, 0,digit));    
                pHead.next  = temp;
                pHead = pHead.next;
                if(0+p.val-10+digit >= 0){
                    digit = 1;
                }
                else{
                    digit= 0;
                }
                p = p.next;
            }
        }
        // 链表1： 5 链表2： 5 这个时候的处理
        if(digit ==1){
            ListNode temp = new ListNode(getSum(0, 0,digit));    
            pHead.next  = temp;
        }

        return newHead;
    }
}

更优美一点的代码

javapublic 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;
    }
}