leetcode 445. Add Two Numbers II

    题目大意：给定两个数，以两个链表的形式给出，算出他们的和，最后返回一个新链表。

　 解题思路：把给定的两个链表逆序，求和，在把求出的和即为所求的链表。

　

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        
        
         if(l1 == null) return l2;
         if(l2 == null) return l1;
         
         int flag = 0;
         ListNode res = new ListNode(0);
         ListNode tmp = new ListNode(0);
         res = tmp;
           
         l1 = reverse(l1);
         l2 = reverse(l2);
           
         while(l1 != null || l2 != null){
             int a = l1 == null?0:l1.val;
             int b = l2 == null?0:l2.val;
             ListNode ss = new ListNode((a+b+flag)%10);
             flag = (a+b+flag) /10;
             tmp.next = ss;
             tmp = ss;
               
             l1 = l1 == null? null:l1.next;
             l2 = l2 == null? null:l2.next;
             
         }
        
         if(flag == 1){
             
             ListNode ss = new ListNode(1);
             tmp.next = ss;
             
         }
            
            
           return reverse(res.next);
            
    }
    
    
    public ListNode reverse(ListNode l1){
        
        if(l1.next == null) return l1;
        
        ListNode p = new ListNode(0);
        ListNode q = new ListNode(0);
        
        p = l1;
        q = l1.next;
        p.next = null;
        
        while(q != null){
            
            p = q;
            q = q.next;
            p.next = l1;
            l1 = p;
            
        }
        
        return l1;
    }
    
    
}