2. Add Two Numbers

Posted 2020-07-30 HenRy

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

/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/

2个数据链相加，向后进位，相当于倒置的加法，我们需要一个中间值来存放进位其他的按位相加即可，在这里遇到一个问题，如何在循环中给个temp指针不断的更新然后又让结果能保持在变量中？

定义了一个Loop和Result，2个Node在一开始都指向同一个地址，然后loop又指向loop复制前的loop.next，第一次也是Result.Next的地址，以此类推，每次重定向都是指向下一级result.next，最后返回result.next即可。

然后第一个和第二个的差别中间变量sum，第一个等于计算了2次sum，而第二个只计算了一次，运行时间比第一次上，因此多次利用的值最好先存储起来。

My submission 1: Accepted  Runtime:234ms

 1 public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
 2          ListNode result=new ListNode(0);
 3             int carray = 0;
 4             ListNode p = l1;
 5             ListNode q = l2;
 6             ListNode loop = result;
 7             while (p != null || q != null)
 8             {
 9                 int a = p != null ? p.val : 0;
10                 int b = q != null ? q.val : 0;
11                
12                 int sum = (a + b + carray)%10;
13 
14                 //loop.val = sum;
15                 //loop = loop.next = new ListNode(0);
16                 loop.next = new ListNode(sum);
17                 loop = loop.next;
18 
19                 carray = (a + b + carray) / 10;
20                 p = p!=null?p.next:null;
21                 q = q!=null?q.next:null;
22                 
23             }
24             if(carray>0)
25                 loop.next=new ListNode(carray);
26             return result.next;
27     }

View Code

My submission 2: Accepted Runtime:209ms

 1  public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
 2          ListNode result=new ListNode(0);
 3             int carray = 0;
 4             ListNode p = l1;
 5             ListNode q = l2;
 6             ListNode loop = result;
 7             while (p != null || q != null)
 8             {
 9                 int a = p != null ? p.val : 0;
10                 int b = q != null ? q.val : 0;
11                
12                 int sum = a + b + carray;
13 
14                 //loop.val = sum;
15                 //loop = loop.next = new ListNode(0);
16                 loop.next = new ListNode(sum%10);
17                 loop = loop.next;
18 
19                 carray = sum / 10;
20                 p = p!=null?p.next:null;
21                 q = q!=null?q.next:null;
22                 
23             }
24             if(carray>0)
25                 loop.next=new ListNode(carray);
26             return result.next;
27     }

View Code

 复杂度：循环的终止取决于2个node的node个数中较长的那一个，因此为O（MAX(m,n)），新的node最长为max(m,n)+1 因此空间复杂度为O(max(m,n)).