No.002 Add Two Numbers

Add Two Numbers

• Total Accepted: 160702
• Total Submissions: 664770
• Difficulty: Medium

　　You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

　　本题的思路其实很简单，两个链表的结构都是从低位到高位的顺序，税后要求返回的链表也是这样的结构。所以我们只需要依次循环两个链表，将对应位的数相加，并判断是否有进位，有进位则将进位累加到下一位即可。

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) { val = x; }
  * }
  */
 public class Num2 {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
         if(l1 == null){
             return l2 ;
         }
         if(l2 == null){
             return l1 ;
         }
         int nextBit = (l1.val + l2.val)/10  ;
         int curBit = (l1.val + l2.val)%10 ;
         ListNode head = new ListNode(curBit) ;
         ListNode temp = head ;
         l1 = l1.next ;
         l2 = l2.next ;
         //当l1、l2对应位均存在时，进行计算
         while(l1 != null && l2 != null){
             curBit = (l1.val + l2.val + nextBit) % 10 ;
             nextBit = (l1.val + l2.val + nextBit)/10 ;
             ListNode node = new ListNode(curBit) ;
             temp.next = node ;
             temp = temp.next ;
             l1 = l1.next ;
             l2 = l2.next ;
         }
         //判断l1是否结束，没有结束继续
         while(l1 != null){
             curBit = (l1.val + nextBit) % 10 ;
             nextBit = (l1.val + nextBit)/10 ;
             ListNode node = new ListNode(curBit) ;
             temp.next = node ;
             temp = temp.next ;
             l1 = l1.next ;
         }
         //判断l2是否结束，没有结束继续
         while(l2 != null){
             curBit = (l2.val + nextBit) % 10 ;
             nextBit = (l2.val + nextBit)/10 ;
             ListNode node = new ListNode(curBit) ;
             temp.next = node ;
             temp = temp.next ;
             l2 = l2.next ;
         }
         //判断最后的进位位是否为0 ，不为0则需要保存下一位
         if(nextBit != 0){
             ListNode node = new ListNode(nextBit) ;
             temp.next = node ;
         }
         return head ;
     }
 }