LeetCode 2. Add Two Numbers题目分析代码

/**
 * 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
        // write your code here
        // save the pre node
        ListNode pre = new ListNode(0);
        // save the now node
        ListNode now = new ListNode(0);
        // create a null node to store the result
        ListNode result = null;
        int val = 0; //
        int add = 0; //jinwei
        
        while( l1 != null || l2 != null )
        {
            val = ((l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + add) % 10;
            add = ((l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + add) / 10;
            l1 = (l1 == null ? null : l1.next);
            l2 = (l2 == null ? null : l2.next);
            now.val = val;
            if(result == null)
            {
                result = now;
            }
            pre = now;
            now = new ListNode(0);
            pre.next = now;
        }
        //最后还要多来一次判断，因为有一种可能，两个链表一样长，最后一位又向上进了一位
        val = ((l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + add) % 10;
        add = ((l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + add) / 10;
        now.val = val;
        //如果最后一位又向上进了一位，新的最后一位不为0，应该保留，否则就为0，应当舍弃
        if(now.val == 0){
            pre.next = null;
        }
        return result;
    }
}