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

class Solution {

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        ListNode c1 = l1;

        ListNode c2 = l2;

        ListNode sentinel = new ListNode(0);

        ListNode d = sentinel; //开始d指向0节点

        int sum = 0;

        while (c1 != null || c2 != null) {

            sum /= 10; //算出进位的数值

            if (c1 != null) {

                sum += c1.val;

                c1 = c1.next;

            }

            if (c2 != null) {

                sum += c2.val;

                c2 = c2.next;

            }

            d.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

            d = d.next;

        }

        

        if (sum / 10 == 1) //考虑最后是否还有进位

            d.next = new ListNode(1);

        

        return sentinel.next;

    }

}

 

转载于:https://www.cnblogs.com/MarkLeeBYR/p/10677738.html