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leetcode2 Add Two Numbers 方法2

思路：凡是对链表的数字的操作，都可以考虑将这些数字转化为一个long或者一个数组(这个思路较好，可以为以后的开发省去好多不必要的步骤)

 

 

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

这个方法就是将链表中的数字串起来，当做一个long，例如2->4->5,可以根据题目具体要求转化成long型的245或542，再做后续的操作，就很容易了。举一反三，链表数字的反序也可以采用这个方法。

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13         public Long listTOLong(ListNode l){
14         long num = 0;
15         long temp =1;
16         int i=0;
17         while(l!=null){
18             num = num+l.val*temp;
19             temp=temp*10;
20             l=l.next;
21         }
22         return num;
23     }   
24     public ListNode longToList(Long num){        
25         ListNode l3 = new ListNode(-1);
26         l3.next = null;
27         ListNode c = l3;
28         c.val=(int)(num%10);
29         num = num/10;
30         while(num>0){        
31             ListNode cnext = new ListNode((int)(num%10));
32             cnext.next=null;
33             c.next=cnext;
34             num = num/10;
35             c=c.next;
36         }
37         return l3;
38     }
39     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
40         if(l1==null&&l2==null){
41             return null;
42         }
43         //链表转long型
44         long num1 = listTOLong(l1);
45         long num2 = listTOLong(l2);
46         //System.out.println("l1:"+num1+" l2:"+num2);
47         long num3 = num1+num2;
48         //System.out.println("l3:"+num3);
49         //long型转链表
50         ListNode l3 = longToList(num3);
51         return l3;
52          
53     }
54 }

 还可以利用结构体的方法

 1 struct ListNode {
 2     int val;
 3     ListNode *next;
 4     ListNode(int x) : val(x), next(NULL) {}
 5 };
 6 
 7 class Solution {
 8 public:
 9     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
10         // Start typing your C/C++ solution below
11         // DO NOT write int main() function
12 //        ListNode *pResult = NULL;
13 //        ListNode **pCur = &pResult;
14 
15         ListNode rootNode(0);
16         ListNode *pCurNode = &rootNode;
17         int a = 0;
18         while (l1 || l2)
19         {
20             int v1 = (l1 ? l1->val : 0);
21             int v2 = (l2 ? l2->val : 0);
22             int temp = v1 + v2 + a;
23             a = temp / 10;
24             ListNode *pNode = new ListNode((temp % 10));
25             pCurNode->next = pNode;
26             pCurNode = pNode;
27             if (l1)
28                 l1 = l1->next;
29             if (l2)
30                 l2 = l2->next;
31         }
32         if (a > 0)
33         {
34             ListNode *pNode = new ListNode(a);
35             pCurNode->next = pNode;
36         }
37         return rootNode.next;
38     }
39 };

 


posted on 2015-06-14 20:24  xqCui  阅读(104)  评论(0)    收藏  举报

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