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2. Add Two Numbers

时间:2019-04-09 16:39:39      阅读:167      评论:0      收藏:0      [点我收藏+]

标签:sts   class   lin   public   lis   repr   sentinel   数值   out   

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;
    }
}

 

2. Add Two Numbers

标签:sts   class   lin   public   lis   repr   sentinel   数值   out   

原文地址:https://www.cnblogs.com/MarkLeeBYR/p/10677738.html

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