/** * 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；