[数据结构与算法]链表求和（面试）

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        /**
        /思路：
        从头到位遍历两链表
        计算每两个节点的和 记下进位
        当一个链表到了尽头 就将另一个未到尽头的链表拼接到答案上
         */
         ListNode ans = new ListNode(0);
         ListNode yummy = ans;
         int flag = 0; 
         int node = 0 ;
         while(l1 !=null && l2 !=null){
             if(l1.val + l2.val + flag >=10){
                 node  =  (l2.val + l1.val +flag)%10 ;
                 flag = 1;
             }else{
                  node =  l2.val + l1.val +flag ;
                 flag = 0 ; 
             }
             ans.next = new ListNode(node);
             ans = ans.next;
             l2 = l2.next;
             l1 = l1.next;
         }
         while(l1 != null ){
             node = l1.val + flag;
            ans.next = new ListNode(node %10);
            flag = node /10;
            ans = ans.next;
            l1 = l1.next;
         }

         while(l2 != null){
              node = l2.val + flag;
            ans.next = new ListNode(node %10);
            flag = node /10;
            ans = ans.next;
            l2 = l2.next;
         }
          if(flag != 0 ){
             ans.next = new ListNode(flag);
         }
          return yummy.next;
    }
}