2. Add Two Numbers

 Add Two Numbers  

Difficulty: Medium

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

public class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}

/*
* 分析：题目中给出两个链表，然后相加，但是一旦相加超过10，就会向下一个节点进位，这里就需要判断如何进行进位
* @titleAddTwoSum.java
* @author taojian
* @date2016-11-17下午1:08:29
*/
public class AddTwoSum {
public static void main(String[] args) {
ListNode l1=new ListNode(2);
l1.next=new ListNode(4);
l1.next.next=new ListNode(3);
ListNode l2=new ListNode(5);
l2.next=new ListNode(6);
l2.next.next=new ListNode(4);
/* while(l1!=null){
System.out.println(l1.val);
//System.out.println(l1.next.val);
l1=l1.next;
}*/
AddTwoSum as=new AddTwoSum();
ListNode sum=as.addTwoNumbers(l1,l2);
while(sum!=null){
System.out.println(sum.val);
sum=sum.next;
}

}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
/* ListNode sum=new ListNode(0);
ListNode list1=l1,list2=l2;
while(l1.next!=null&&l2.next!=null){
sum.val=l1.val+l2.val;
l1=l1.next;
l2=l2.next;
sum=sum.next;
//以上题目没看清
}
return sum;*/
//设置头节点，数据为0
ListNode head = new ListNode(0);
//两个链表p,q,一个新的链表currsum用来存放p,q节点相加
ListNode list1 = l1, list2 = l2, currsum = head;
int m = 0;
while (list1 != null || list2 != null) {
//由于链表的长度不一致，当某一个链表空的时候，应该把0置位
int x,y;
if(list1!=null) x=list1.val;
else x=0;
if(list2!=null) y=list2.val;
else y=0;
//如果过了10，就会进位
int sum = m + x + y;
m = sum / 10;
//超过10的数取各位数
currsum.next = new ListNode(sum % 10);
currsum = currsum.next;
if (list1 != null)
list1 = list1.next;
if (list2!= null)
list2 = list2.next;
}//这个需要判断

  if (m > 0) {
         currsum.next = new ListNode(m);
     }return head.next;
}
}