leetcode|leetcode_2 leetcode|leetcode

You are given two non-empty linked lists representing two non-negative integers. 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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.

【leetcode|leetcode_2】第一次的写法，过于复杂繁琐，考虑了太多因素，也导致了很多代码的重读

class Solution { public: ListNode* addTwoNumbers(ListNode* l1,ListNode* l2) { if (l1==NULL||l2==NULL) { return NULL; } ListNode* ptr1 = l1; ListNode* ptr2 = l2; ListNode* l3 = new ListNode(0); ListNode* ptr3 = l3; while(ptr1!=NULL && ptr2!=NULL) { ptr3->val=ptr1->val+ptr2->val; if(ptr3->val >= 10) { ptr3->val = ptr3->val%10; ptr3->next = new ListNode(1); ptr3 = ptr3->next; if(ptr1->next==NULL && ptr2->next==NULL) { ptr1 = ptr1->next; ptr2 = ptr2->next; } else if(ptr1->next==NULL && ptr2->next!=NULL) { ptr1->next=new ListNode(0); ptr1 = ptr1->next; ptr2 = ptr2->next; ptr2->val+=1; } else if(ptr2->next==NULL && ptr1->next!=NULL) { ptr2->next=new ListNode(0); ptr2 = ptr2->next; ptr1 = ptr1->next; ptr1->val+=1; } else if (ptr1->next!=NULL && ptr2->next!=NULL) { ptr1 = ptr1->next; ptr1->val+=1; ptr2 = ptr2->next; } } else { ptr3->next = new ListNode(0); if(ptr1->next==NULL && ptr2->next==NULL) { ptr1 = ptr1->next; ptr2 = ptr2->next; delete ptr3->next; ptr3->next=NULL; break; } else if(ptr1->next==NULL && ptr2->next!=NULL) { ptr1->next=new ListNode(0); ptr1 = ptr1->next; ptr2 = ptr2->next; }else if(ptr2->next==NULL && ptr1->next!=NULL) { ptr2->next=new ListNode(0); ptr2 = ptr2->next; ptr1 = ptr1->next; }else if (ptr1->next!=NULL && ptr2->next!=NULL) { ptr1 = ptr1->next; ptr2 = ptr2->next; } ptr3 = ptr3->next; } } return l3; } };

第二次换了一下思路，不管长度是否一致，都在短的数字后面加上一个零，方便与长的数字进行后续相加，也要判断后面是否为NULL

class Solution { public: ListNode* addTwoNumbers(ListNode* l1,ListNode* l2) { if (l1==NULL||l2==NULL) { return NULL; } ListNode* ptr1 = l1; ListNode* ptr2 = l2; ListNode* l3 = new ListNode(0); ListNode* ptr3 = l3; while(ptr3!=NULL) { ptr3->val=ptr1->val+ptr2->val; if(ptr1->next==NULL && ptr2->next!=NULL) ptr1->next=new ListNode(0); else if(ptr2->next==NULL && ptr1->next!=NULL) ptr2->next=new ListNode(0); else if(ptr1->next==NULL&&ptr2->next==NULL) { if(ptr3->val==0) break; ptr1->next=new ListNode(0); ptr2->next=new ListNode(0); } ptr1 = ptr1->next; ptr2 = ptr2->next; ptr1->val = ptr1->val + ptr3->val/10; ptr3->val = ptr3->val%10; if(ptr1->val+ptr2->val != 0) ptr3->next = new ListNode(0); else if(ptr1->next!=NULL || ptr2->next!=NULL) ptr3->next = new ListNode(0); ptr3 = ptr3->next; } return l3; } };