算法步步为营(02)-两数之和

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = null, tail = null;
        // 将 carry 初始化为 0
        int carry = 0;
        while (l1 != null || l2 != null) {
            int n1 = l1 != null ? l1.val : 0;
            int n2 = l2 != null ? l2.val : 0;
            int sum = n1 + n2 + carry;
            if (head == null) {
                head = tail = new ListNode(sum % 10);
            } else {
                tail.next = new ListNode(sum % 10);
                tail = tail.next;
            }
            carry = sum / 10;
            if (l1 != null) {
                l1 = l1.next;
            }
            if (l2 != null) {
                l2 = l2.next;
            }
        }
        if (carry > 0) {
            tail.next = new ListNode(carry);
        }
        return head;
    }
}

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        return add(l1, l2, 0);
    }

    /**
        返回两个链表相加的头部
     */
    public ListNode add(ListNode l1, ListNode l2, int bit) {
        if (l1 == null && l2 == null && bit == 0) {
            return null;
        }
        int val = bit;
        if (l1 != null) {
            val += l1.val;
            l1 = l1.next;
        }
        if (l2 != null) {
            val += l2.val;
            l2 = l2.next;
        }
        ListNode node = new ListNode(val % 10);
        node.next = add(l1, l2, val / 10);
        return node;
    }
}