[Lintcode] #167 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
ListNode newHead = new ListNode(-1);
ListNode pre = newHead;
int carry = 0;
while (l1 != null || l2 != null) {
int sum = carry;
if (l1 != null) {
sum += l1.val;
l1 = l1.next;
}
if (l2 != null) {
sum += l2.val;
l2 = l2.next;
}

if (sum > 9) {
carry = 1;
sum -= 10;
} else
carry = 0;
ListNode cur = new ListNode(sum);
pre.next = cur;
pre = cur;
}
if (carry == 1)
pre.next = new ListNode(1);
return newHead.next;
}
}