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You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
思路：从尾巴开始加数，和大于10就变carry和值。需要注意当一个list读完后，那个list的值就是0，只需要加另一个list。最后判断carry的值，如果是0就直接返回，是1的话还要补上一个node。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode dummy=new ListNode(-1); ListNode copy=dummy; int carry=0; while(l1!=null||l2!=null) { int va1=0; if(l1!=null) { va1=l1.val; l1=l1.next; } int va2=0; if(l2!=null) { va2=l2.val; l2=l2.next; } if((va1+va2+carry)>=10) { copy.next=new ListNode(va1+va2+carry-10); carry=1; } else { copy.next=new ListNode(va1+va2+carry); carry=0; } copy=copy.next; } if(carry==0) { copy.next=null; } else { copy.next=new ListNode(1); } return dummy.next; } }
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原文地址：http://www.cnblogs.com/Machelsky/p/5894761.html
