[LeetCode-Java]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

解：由于int大小限制，不可转化为int相加处理。对应位相加的过程中需注意末尾进位情况。

public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode result = l1;        boolean flag = false;        while(true){            //对应位相加  考虑前一位的进位flag            l1.val = l1.val + l2.val + (flag?1:0);            //根据相加的结果  设置进位标志            if (l1.val >= 10){                l1.val = l1.val - 10;                flag = true;            }else {                flag = false;            }            if (l1.next == null || l2.next == null)  break;            l1 = l1.next;            l2 = l2.next;        }        //l2长度比较长，作相应处理        if (l1.next == null && l2.next != null){            l2 = l2.next;            l1.next = l2;           dealWithEnd(l2,flag);        }        //l1的长度比较长        if (l2.next == null && l1.next != null){            l1 = l1.next;            dealWithEnd(l1,flag);        }        //两者长度相同，需根据进位情况增加一个最高位        if (l1.next == null && l2.next == null){            if (flag) {                ListNode end = new ListNode(1);                l1.next = end;            }        }        return result;    }    //对附带进位信息的链表进行处理    static void dealWithEnd(ListNode listNode,Boolean flag){        while (true){            listNode.val  += (flag?1:0);            if (listNode.val >= 10){                listNode.val = listNode.val - 10;                flag = true;            }else {                flag = false;                break;            }            if (listNode.next == null){                if (flag) {                    ListNode end = new ListNode(1);                    listNode.next = end;                }                break;            }            listNode = listNode.next;        }    }