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2. Add Two Numbers

1. 问题描述

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

Tags: Linked List Math
Similar Problems: (M) Multiply Strings (E) Add Binary

2. 解答思路

3. 代码

  1 /**
  2 * Definition for singly-linked list.
  3 * struct ListNode {
  4 *     int val;
  5 *     ListNode *next;
  6 *     ListNode(int x) : val(x), next(NULL) {}
  7 * };
  8 */
  9 #include <iostream>
 10 using namespace std;
 11 struct ListNode {
 12     int val;
 13     ListNode *next;
 14     ListNode(int x) : val(x), next(NULL) {}
 15 };
 16 class Solution {
 17 public:
 18     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
 19         if (NULL == l1)
 20         {
 21             return l2;
 22         }
 23         if (NULL == l2)
 24         {
 25             return l1;
 26         }
 27         return addTwoNum(l1, l2);
 28     }
 29 public:
 30     void PrintList(ListNode* l1)
 31     {
 32         if (NULL == l1)
 33         {
 34             return;
 35         }
 36         ListNode *pt = l1;
 37         while (NULL != pt)
 38         {
 39             std::cout<< pt->val << " ";
 40             pt = pt->next;
 41         }
 42 
 43     }
 44 private:
 45     ListNode* addTwoNum(ListNode* ptLar, ListNode* ptSma)
 46     {
 47         ListNode *pLar = ptLar;
 48         ListNode *pSma = ptSma;
 49         int t = 0;
 50         while (NULL != pSma && NULL != pLar)
 51         {
 52             int sum = pLar->val + pSma->val + t;
 53             pLar->val = sum%10;
 54             pSma->val = pLar->val;
 55             t = sum/10;
 56             if ((NULL != pSma->next && NULL != pLar->next))
 57             {
 58                 pSma = pSma->next;
 59                 pLar = pLar->next;
 60             }
 61             else
 62             {
 63                 break;
 64             }
 65 
 66         }
 67 
 68         if (0 == t)
 69         {
 70             if (NULL == pSma->next)
 71             {
 72                 return ptLar;
 73             }
 74             else
 75             {
 76                 return ptSma;
 77             }
 78         }
 79 
 80         if (NULL == pSma->next && NULL == pLar->next)
 81         {
 82             ListNode *pNew = new ListNode(1);
 83             pLar->next = pNew;
 84             return ptLar;
 85         }
 86 
 87         if ( NULL != pLar->next)
 88         {
 89             addone(pLar);
 90             return ptLar;
 91         }
 92         else
 93         {
 94             addone(pSma);
 95             return ptSma;
 96         }
 97 
 98     }
 99     void addone(ListNode *pLar, int t = 1)
100     {
101         pLar = pLar->next;
102         while (NULL != pLar)
103         {
104             int sum = pLar->val + t;
105             pLar->val = sum%10;
106             t = sum/10;
107             if (0 == t)
108             {
109                 break;                
110             }
111             else
112             {
113                 if (NULL == pLar->next)
114                 {
115                     ListNode *pNew = new ListNode(1);
116                     pLar->next = pNew;
117                     break;
118                 }
119                 else
120                 {
121                     pLar = pLar->next;
122                 }
123             }
124         }
125     }
126 };

4. 反思

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原文地址：https://www.cnblogs.com/whl2012/p/5582235.html