Lincode56 - Add Two Numbers - easy

/**
 * 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

        int digit1 = l1.val;
        int digit2 = l2.val;
        int digitS = (digit1 + digit2) % 10;
        int c = (digit1 + digit2) / 10;
        l1 = l1.next;
        l2 = l2.next;

        ListNode sum = new ListNode(digitS);
        ListNode head = sum;

        while (!(l1 == null && l2 == null && c == 0)){
            digit1 = l1 == null ? 0 : l1.val;
            digit2 = l2 == null ? 0 : l2.val;
            digitS = (digit1 + digit2 + c) % 10;
            c = (digit1 + digit2 + c) / 10;
            l1 = l1 == null ? l1 : l1.next;
            l2 = l2 == null ? l2 : l2.next;
            sum.next = new ListNode(digitS);
            sum = sum.next;
        }

        return head;
    }
}

1. 小心处理l1和l2其中有一个先null了的情况，可以digit1 = l1 == null ? 0 : l1.val;
2. 还有小心l1后移的情况，如果l1已经是null了，那就不能后移了，所以要l1 = l1 == null ? l1 : l1.next;
3. 小心后面每次加的时候要多加个c；