LeetCode2. 两数相加

☆☆☆解法：考虑进位值 以及 较短的链表补0

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummyHead = new ListNode(0);
        ListNode cur = dummyHead;
        int carry = 0; // 上一位的进位值
        while (l1 != null || l2 != null) {
            // 如果一个链表较短则在前面补0
            int val1 = l1 == null ? 0 : l1.val;
            int val2 = l2 == null ? 0 : l2.val;
            int temp = val1 + val2 + carry;
            carry = temp / 10;

            cur.next = new ListNode(temp % 10);
            cur = cur.next;
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
        }
        // 如果两个链表全部遍历完毕后，进位值为 1，则在新链表最前方添加节点 1
        if (carry > 0) { // carry == 1
            cur.next = new ListNode(carry);
        }
        return dummyHead.next;
    }
}