LeetCode2 Add Two Numbers

题意：

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. (Medium)

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

分析：

链表操作题，算法没什么复杂的，注意语言上不出问题就行，其次对于写法要尽可能简练，自己第一种写法就很冗长，改进后得到方法2。

链表注意：空表判断，dummy node应用等

代码1：

 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 class Solution {
10 public:
11     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
12         if (l1 == nullptr) {
13             return l2;
14         }
15         if (l2 == nullptr) {
16             return l1;
17         }
18         ListNode dummy(0);
19         ListNode* result;
20         dummy.next = result;
21         result = &dummy;
22         int carry = 0;
23         while (l1 != nullptr && l2 != nullptr ) {
24             ListNode* newNode = new ListNode(0);
25             result -> next = newNode;
26             int temp = l1 -> val + l2 -> val + carry;
27             if (temp >= 10){
28                 carry = 1;
29                 newNode -> val = temp - 10;
30             }
31             else{
32                 carry = 0;
33                 newNode -> val = temp;
34             }
35             l1 = l1 -> next;
36             l2 = l2 -> next;
37             result = newNode;
38         }
39         
40         if (l1 != nullptr) {
41             while (l1 != nullptr) {
42                 ListNode* newNode = new ListNode(0);
43                 //仍然有可能进位！
44                 int temp = carry + l1 -> val;
45                 if (temp >= 10){
46                     carry = 1;
47                     newNode -> val = temp - 10;
48                 }
49                 else{
50                     carry = 0;
51                     newNode -> val = temp;
52                 }
53                 result -> next = newNode;
54                 result = newNode;
55                 l1 = l1 -> next;
56             }
57         }
58         if (l2 != nullptr) {
59             while (l2 != nullptr) {
60                 ListNode* newNode = new ListNode(0);
61                 int temp = carry + l2 -> val;
62                 if (temp >= 10){
63                     carry = 1;
64                     newNode -> val = temp - 10;
65                 }
66                 else{
67                     carry = 0;
68                     newNode -> val = temp;
69                 }
70                 result -> next = newNode;
71                 result = newNode;
72                 l2 = l2 -> next;
73             }
74         }
75         if(carry == 1){
76              ListNode* newNode = new ListNode(0);
77              newNode -> val = 1;
78              result -> next = newNode;
79         }
80         return dummy.next;
81     }
82 };

各种条件判定过于繁琐，代码又臭又长...

改进后将l1,l2,carry的判断放在一起处理，同时用/ 和 %代替if语句得到代码2

代码2：

 1 class Solution {
 2 public:
 3     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
 4         ListNode dummy(0);
 5         ListNode* result = & dummy;
 6         int carry = 0;
 7         while (l1 != nullptr || l2 != nullptr || carry == 1 ) {
 8             result -> next = new ListNode(0);
 9             result = result -> next;
10             int tempVal = 0;
11             if (l1 != nullptr) {
12                 tempVal += l1 -> val;
13                 l1 = l1 -> next;
14             }
15             if (l2 != nullptr) {
16                 tempVal += l2 -> val;
17                 l2 = l2 -> next;
18             }
19             if (carry == 1) {
20                 tempVal++;
21             }
22             //不用if判断
23             result -> val = tempVal % 10;
24             carry = tempVal / 10;
25         }
26         return dummy.next;
27     }
28 };

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原文地址：https://www.cnblogs.com/wangxiaobao/p/5723681.html