leetcode刷题： 002 Add Two Numbers

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

首先使用链表实现等长无进位的求和，即实现 (1 —> 2 -> 3) + (1 -> 2 -> 3)=（2 -> 3 -> 6）

 #include <iostream>

 using namespace std;

 struct ListNode {
      int val;
      ListNode *next;
      ListNode(int x) : val(x), next(NULL) {}
 };

 class Solution {
 public:
     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {

         int val1 = , val2 = ;

         ListNode *l = NULL;
         ListNode *l_end = NULL;
         ListNode *l_new = NULL;

         val1 = l1->val;
         val2 = l2->val;
         l1 = l1->next;
         l2 = l2->next;
         l_new = (ListNode *)new ListNode((val1 + val2) % );
         l_new->next = NULL;
         l = l_new;
         l_end = l;
         while (l1 != NULL || l2 != NULL){

                 val1 = l1->val;
                 val2 = l2->val;
                 l1 = l1->next;
                 l2 = l2->next;

             l_new = (ListNode *)new ListNode((val1 + val2)%);
             l_new->next = NULL;

             l_end->next = l_new;
             l_end = l_new;
         }

         return l;

     }
 };

 int main()
 {
     ListNode *l1, *l2;
     ListNode *l;
     l1 = (ListNode *)new ListNode();
     l2 = (ListNode *)new ListNode();
     l1->next = (ListNode *)new ListNode();
     l2->next = (ListNode *)new ListNode();
     l1->next->next = (ListNode *)new ListNode();
     l2->next->next = (ListNode *)new ListNode();
     Solution s;
     l = s.addTwoNumbers(l1, l2);
     while (l != NULL){
         cout << l->val << endl;
         l = l->next;
     }
     while ();
 }

运行结果：

然后实现不等长无进位的求和，即实现 (1 —> 2 -> 3) + (1)=（2 -> 2 -> 3）

 #include <iostream>

 using namespace std;

 struct ListNode {
      int val;
      ListNode *next;
      ListNode(int x) : val(x), next(NULL) {}
 };

 class Solution {
 public:
     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {

         int val1 = , val2 = ;

         ListNode *l = NULL;
         ListNode *l_end = NULL;
         ListNode *l_new = NULL;

         val1 = l1->val;
         val2 = l2->val;
         l1 = l1->next;
         l2 = l2->next;
         l_new = (ListNode *)new ListNode((val1 + val2) % );
         l_new->next = NULL;
         l = l_new;
         l_end = l;
         while (l1 != NULL || l2 != NULL){
             if (l1 == NULL){
                 val2 = l2->val;
                 l2 = l2->next;
                 val1 = ;
             }
             else if (l2 == NULL){
                 val1 = l1->val;
                 l1 = l1->next;
                 val2 = ;
             }
             else{
                 val1 = l1->val;
                 val2 = l2->val;
                 l1 = l1->next;
                 l2 = l2->next;
             }

             l_new = (ListNode *)new ListNode((val1 + val2)%);
             l_new->next = NULL;

             l_end->next = l_new;
             l_end = l_new;
         }

         return l;

     }
 };

 int main()
 {
     ListNode *l1, *l2;
     ListNode *l;
     l1 = (ListNode *)new ListNode();
     l2 = (ListNode *)new ListNode();
     l1->next = (ListNode *)new ListNode();
     //l2->next = (ListNode *)new ListNode(2);
     l1->next->next = (ListNode *)new ListNode();
     //l2->next->next = (ListNode *)new ListNode(3);
     Solution s;
     l = s.addTwoNumbers(l1, l2);
     while (l != NULL){
         cout << l->val << endl;
         l = l->next;
     }
     while ();
 }

运行结果：

最后实现不等长有进位的求和，即实现题目要求（注意最后一位有进位的情况）

 #include <iostream>

 using namespace std;

 struct ListNode {
      int val;
      ListNode *next;
      ListNode(int x) : val(x), next(NULL) {}
 };

 class Solution {
 public:
     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {

         int val1 = , val2 = , carry = ;

         ListNode *l = NULL;
         ListNode *l_end = NULL;
         ListNode *l_new = NULL;

         val1 = l1->val;
         val2 = l2->val;
         l1 = l1->next;
         l2 = l2->next;
         l_new = (ListNode *)new ListNode((val1 + val2 + carry) % );
         l_new->next = NULL;
         l = l_new;
         carry = (val1 + val2 + carry) / ;
         l_end = l;
         while (l1 != NULL || l2 != NULL){
             if (l1 == NULL){
                 val2 = l2->val;
                 l2 = l2->next;
                 val1 = ;
             }
             else if (l2 == NULL){
                 val1 = l1->val;
                 l1 = l1->next;
                 val2 = ;
             }
             else{
                 val1 = l1->val;
                 val2 = l2->val;
                 l1 = l1->next;
                 l2 = l2->next;
             }

             l_new = (ListNode *)new ListNode((val1 + val2 + carry)%);
             l_new->next = NULL;
             carry = (val1 + val2 + carry) / ;

             l_end->next = l_new;
             l_end = l_new;
         }
         if (carry != ){
             l_new = (ListNode *)new ListNode(carry);
             l_new->next = NULL;
             l_end->next = l_new;
             l_end = l_new;
         }
         return l;

     }
 };

 int main()
 {
     ListNode *l1, *l2;
     ListNode *l;
     l1 = (ListNode *)new ListNode();
     l2 = (ListNode *)new ListNode();
     l1->next = (ListNode *)new ListNode();
     l2->next = (ListNode *)new ListNode();
     l1->next->next = (ListNode *)new ListNode();
     l2->next->next = (ListNode *)new ListNode();
     Solution s;
     l = s.addTwoNumbers(l1, l2);
     while (l != NULL){
         cout << l->val << endl;
         l = l->next;
     }
     while ();
 }

运行结果：

因为是一边学C++，一边刷leetcode，所以有什么问题，十分感谢您能指点。