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
这是一个简单的数字相加,主要边界条件和进位就行
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode res = new ListNode(0);
ListNode head=res;
int flag=0;
int carry=0;
while (l1!=null || l2!=null)
{
int s=0;
if(l1!=null)
{
s=l1.val;
l1=l1.next;
}
if(l2!=null)
{
s+=l2.val;
l2=l2.next;
}
s+=carry;
if(flag==0)
{
res.val=s%10;
flag=1;
}
else
{
res.next=new ListNode(s%10);
res=res.next;
}
if(s>=10)
{
carry=1;
}
else
{
carry=0;
}
}
if(carry==1)
{
res.next=new ListNode(1);
}
return head;
}
}
这个问题并不是很难,注意一些边界条件还是很容易的
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原文地址:http://blog.csdn.net/jianxia_wzx/article/details/45825089