[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

public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode ln1, head; int flag = 0; head = ln1 = new ListNode(0); while (l1 != null || l2 != null) { if (l1 != null && l2 != null) { if (l1.val + l2.val + flag >= 10) { ln1.val = l1.val + l2.val + flag - 10; flag = 1; } else { ln1.val = l1.val + l2.val + flag; flag = 0; } l1 = l1.next; l2 = l2.next; } else if (l1 == null) { if (l2.val + flag >= 10) { ln1.val = l2.val + flag - 10; flag = 1; } else { ln1.val = l2.val + flag; flag = 0; } l2 = l2.next; } else if (l2 == null) { if (l1.val + flag >= 10) { ln1.val = l1.val + flag - 10; flag = 1; } else { ln1.val = l1.val + flag; flag = 0; } l1 = l1.next; } if (l1 != null || l2 != null) ln1.next = new ListNode(0); if (flag == 1) { ln1.next = new ListNode(0); ln1.next.val = 1; } ln1 = ln1.next; } return head; }

这是我写的，虽然通过了但是我觉得还可以提高，点了一下看到了一个别人写程序，亮瞎我眼了。

/** * 很简便的一个写法，至少甩我两条街 * @param l1 * @param l2 * @return */ public ListNode addTwoNumbers2(ListNode l1, ListNode l2) { ListNode dummyHead = new ListNode(0); ListNode p = l1, q = l2, curr = dummyHead; int carry = 0; while (p != null || q != null) { int x = (p != null) ? p.val : 0; int y = (q != null) ? q.val : 0; int sum = carry + x + y; carry = sum / 10; curr.next = new ListNode(sum % 10); curr = curr.next; if (p != null) p = p.next; if (q != null) q = q.next; } if (carry > 0) { curr.next = new ListNode(carry); } return dummyHead.next; }