LeetCode：2. 两数相加

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        int chu = (l1.val+l2.val)/10;  //记录每次相加后的进位数
        ListNode curr = new ListNode((l1.val+l2.val)%10,null); //当前节点

        ListNode root = curr;//根节点
        //同时满足三个条件，循环才会结束
        while (l1.next!=null || l2.next!=null || chu!=0){ 

            //如果节点已经是空，循环未结束，则val设为0；
            int l1Val = 0;
            if(l1.next != null){
                l1Val = l1.next.val;
                //不为空则继续下一次
                l1 = l1.next;
            }

            int l2Val = 0;
            if(l2.next != null){
                l2Val = l2.next.val;
                //不为空则继续下一次
                l2 = l2.next;
            }

            //计算三者的和
            int he = l1Val + l2Val + chu;
            //重置进位数
            chu = he/10;
            //建立节点
            ListNode listNode = new ListNode(he%10,null);
            //指向建立的节点
            curr.next = listNode;

            //下一个循环
            curr = listNode;
            
        }

        return root;
    }
}