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
思路：模拟题，从头向后依次遍历两个链表相加。
注意：1. 用dummy head简便处理。2. 注意carry的处理，尤其最高位进位的情况。3. 注意引用null的判断。
public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { // printList(l1); // printList(l2); ListNode p = new ListNode(-1); ListNode head = p; int c = 0; while (l1 != null || l2 != null || c == 1) { int curInt = 0; if (l1 != null) curInt += l1.val; if (l2 != null) curInt += l2.val; curInt += c; if (curInt > 9) { c = 1; curInt -= 10; } else c = 0; p.next = new ListNode(curInt); p = p.next; if (l1 != null) l1 = l1.next; if (l2 != null) l2 = l2.next; } // printList(head.next); return head.next; } void printList(ListNode node) { while (node != null) { System.out.print(node.val); if (node.next != null) System.out.print("->"); node = node.next; } System.out.println(); } public static void main(String[] args) { ListNode l1 = new ListNode(2); l1.next = new ListNode(4); l1.next.next = new ListNode(5); ListNode l2 = new ListNode(5); l2.next = new ListNode(6); l2.next.next = new ListNode(4); new Solution().addTwoNumbers(l1, l2); }}class ListNode { int val; ListNode next; ListNode(int x) { val = x; next = null; }}