[LeetCode]Add Two Numbers

Question：

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

先贴上参考答案：

public ListNode addTwoNumbers(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;
}

做本题问题所在：

1、虽然写出了前提条件/不变式/终止条件/边界条件，但是没有形成完整的流程图，导致不能一次coding success；

应该形成以下流程图为框架：



2、应该使用三元操作符，简化代码阅读：

int x = (p != null) ? p.val : 0;

3、应该以p、q==null为边界条件，而不以p.next、q.next==null

4、让dummyHead 指向了一个头结点，但是最后返回时

return dummyHead.next;

，显然更为高明。