每日一题7

题目：2. 两数相加

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = null, tail = null;
        int carry = 0;
        while (l1 != null || l2 != null) {   
            int n1 = l1 != null ? l1.val : 0;     //如果后续节点为空，则赋值为0，如果不这样处理会报错
            int n2 = l2 != null ? l2.val : 0;
            int sum = n1 + n2 + carry;           
            if (head == null) {
                head = tail = new ListNode(sum % 10);   // 解决了首节点为0的问题
            } else {
                tail.next = new ListNode(sum % 10);   // tail指针遍历整个链表
                tail = tail.next;
            }
            carry = sum / 10;  
            if (l1 != null) {
                l1 = l1.next;    //l1链表遍历
            }
            if (l2 != null) {
                l2 = l2.next;   //l2链表遍历
            }
        }
        if (carry > 0) {    //如果最后一个节点的和大于10，则对应产生的进位要多加一个节点
            tail.next = new ListNode(carry);
        }
        return head;
    }
}