2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

链表存的是数字的反序，实际是342+465=807

 

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode cur = dummy; //开始d指向0节点
    int sum = 0;
    while (l1 != null || l2 != null) {
        sum /= 10; //算出进位的数值
        if (l1 != null) {
            sum += l1.val;
            l1 = l1.next;
        }
        if (l2 != null) {
            sum += l2.val;
            l2 = l2.next;
        }
        cur.next = new ListNode(sum % 10);//如4+8=12则把2存入下一个结点，进位的1在循环开始地方算出来
        cur= cur.next;
    }

    if (sum / 10 == 1) //考虑最后是否还有进位
        cur.next = new ListNode(1);

    return dummy.next;
}