Leetcode: 2. Add Two Numbers

一直想用一个进位变量来存贮进位值，但老是考虑不周全，下面是我自己写的bug代码，考虑不周，因为l1或者l2都有可能为null

class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode listEnd = new ListNode(0);ListNode list = listEnd;int carry = 0;while (l1 != null || l2 != null) {int sum = 0;if(l1!=null) {sum =carry+l1.val;l1 = l1.next;}if(l2!=null) {sum =l2.val+sum;l2 = l2.next;}list.next = new ListNode(sum % 10);carry = sum/10;list = list.next;}if (carry== 1)list.next = new ListNode(1);return listEnd.next;}
}

acccpt：https://discuss.leetcode.com/topic/799/is-this-algorithm-optimal-or-what

import java.util.Map;public class Test2 {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode listEnd = new ListNode(0);   //作为最后返回的链表，返回的是nextListNode list = listEnd;  //一直在遍历的链表,因为list指针一直是往后移动，所以只用一个指针是完成不了任务的，必须有一个作为缓存int sum = 0;while (l1 != null || l2 != null) {sum =sum/10;     //求进位，这里就只能用这种方式。我一直在想用一个变量来存储，但往往会考虑不周if(l1!=null) {sum =sum+l1.val;l1 = l1.next;}if(l2!=null) {sum =sum+l2.val;l2 = l2.next;}list.next = new ListNode(sum % 10);    //动态产生链表节点
             list = list.next;}if (sum / 10 == 1)      //最后的和大于10的话，则进位list.next = new ListNode(1);return listEnd.next;}
}

转载于:https://www.cnblogs.com/Michael2397/p/8025059.html

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