链表求和

相反

3->1->5->null

5->9->2->null

8->0->8->null

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
public ListNode addLists(ListNode l1, ListNode l2) {
// write your code here
if(l1 == null)  return l2;
if(l2 == null)  return l1;

ListNode result = null;
ListNode pNode = null;
ListNode pNext = null;
ListNode p = l1;
ListNode q = l2;
int up=0;
while(p!=null&&q!= null){
pNext = new ListNode(p.val+q.val +up);
up = pNext.val/10;
pNext.val = pNext.val%10;

if(result == null){
result = pNode = pNext;
}
else{
pNode.next = pNext;
pNode = pNext;
}
p = p.next;
q = q.next;
}

while(p!= null){
pNext = new ListNode(p.val + up);
up = pNext.val/10;
pNext.val = pNext.val%10;
pNode.next = pNext;
pNode = pNext;
p = p.next;
}
while(q!= null){
pNext = new ListNode(q.val + up);
up = pNext.val/10;
pNext.val = pNext.val%10;
pNode.next = pNext;
pNode = pNext;
q = q.next;
}
if(up!= 0){
pNext = new ListNode(up);
pNode.next = pNext;
}
return result;
}
}