LeetCode(2)--Add Two Numbers

题目如下：
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

解题思路：这里实现的代码比较繁琐，使用了二表合并的思想，先对两个表等长的部分加和，然后再对较长的表进行操作。

提交代码：

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        if(l1 == null || l2 == null){            System.err.println("list node is null!");            return null;        }        int flag = 0; // 标识，是否进位        ListNode node = null;          // 操作表        int value1 = l1.val;        int value2 = l2.val;        int value = value2 + value1;  //对两个表同位的数求和        if(value < 10){            node = new ListNode(value);        }else{            node = new ListNode(value - 10);            flag = 1;        }        ListNode head = node;   //表头，需要返回的表        while(true){ // 对两表同位的数进行操作            if(l1.next == null || l2.next == null){                break;            }            l1 = l1.next;            l2 = l2.next;            value1 = l1.val;            value2 = l2.val;            if(flag == 0){                 value = value1 + value2;            }else{                value = value1 + value2 + 1;            }            if(value < 10){                node.next = new ListNode(value);                flag = 0;            }else{                node.next = new ListNode(value - 10);                flag = 1;            }            node = node.next;        }        while(l1.next != null){ //如果l1较长            l1 = l1.next;            if(flag == 1){                if(l1.val + 1 < 10){                    node.next = new ListNode(l1.val + 1);                    flag = 0;                }else{                    node.next = new ListNode(l1.val + 1 - 10);                    flag = 1;                }            }else{                node.next = new ListNode(l1.val);            }            node = node.next;        }        while(l2.next != null){ //如果l2较长            l2 = l2.next;            if(flag == 1){                if(l2.val + 1 < 10){                    node.next = new ListNode(l2.val + 1);                    flag = 0;                }else{                    node.next = new ListNode(l2.val + 1 - 10);                    flag = 1;                }            }else{                node.next = new ListNode(l2.val);            }            node = node.next;        }        if(flag == 1){  //如果有进位，补上最高位            node.next = new ListNode(1);            node = node.next;        }            node.next = null;        return head;    }