Leetcode2 ADD Numbers

Test instructions

You are given, linked lists representing, and non-negative numbers. The digits is stored in reverse order and all of their nodes contain a single digit. ADD the numbers and return it as a linked list.

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

Analysis:

List operation problem, the algorithm is not complex, pay attention to the language is not a problem on the line, followed by the wording to be as concise as possible, his first kind of writing is very lengthy, improved to get method 2.

List Note: Empty table judgment, dummy node application, etc.

Code Listing 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 classSolution {Ten  Public: Onelistnode* addtwonumbers (listnode* L1, listnode*L2) { A         if(L1 = =nullptr) { -             return< -         } the         if(L2 = =nullptr) { -             returnL1; -         } -ListNode Dummy (0); +listnode*result; -Dummy.next =result; +result = &dummy; A         intcarry =0; at          while(L1! = nullptr && L2! =nullptr) { -listnode* NewNode =NewListNode (0); -Result-Next =NewNode; -             inttemp = L1 val + L2 Val +carry; -             if(Temp >=Ten){ -carry =1; inNewNode, val = temp-Ten; -             } to             Else{ +carry =0; -NewNode, val =temp; the             } *L1 = L1Next; $L2 = L2Next;Panax Notoginsengresult =NewNode; -         } the          +         if(L1! =nullptr) { A              while(L1! =nullptr) { thelistnode* NewNode =NewListNode (0); +                 //there is still a possibility of rounding!  -                 inttemp = carry + L1Val; $                 if(Temp >=Ten){ $carry =1; -NewNode, val = temp-Ten; -                 } the                 Else{ -carry =0;WuyiNewNode, val =temp; the                 } -Result-Next =NewNode; Wuresult =NewNode; -L1 = L1Next; About             } $         } -         if(L2! =nullptr) { -              while(L2! =nullptr) { -listnode* NewNode =NewListNode (0); A                 inttemp = carry + L2Val; +                 if(Temp >=Ten){ thecarry =1; -NewNode, val = temp-Ten; $                 } the                 Else{ thecarry =0; theNewNode, val =temp; the                 } -Result-Next =NewNode; inresult =NewNode; theL2 = L2Next; the             } About         } the         if(Carry = =1){ thelistnode* NewNode =NewListNode (0); theNewNode, val =1; +Result-Next =NewNode; -         } the         returnDummy.next;Bayi     } the};

Various conditions are too cumbersome to determine, the code is smelly and long ...

After the improvement , the L1,l2,carry judgment is processed together, and the/and% are substituted for the IF statement to get the code 2.

Code Listing 2:

1 classSolution {2  Public:3listnode* addtwonumbers (listnode* L1, listnode*L2) {4ListNode Dummy (0);5listnode* result = &dummy;6         intcarry =0;7          while(L1! = nullptr | | L2! = NULLPTR | | carry =1 ) {8Result-Next =NewListNode (0);9result = ResultNext;Ten             intTempVal =0; One             if(L1! =nullptr) { ATempVal + = L1Val; -L1 = L1Next; -             } the             if(L2! =nullptr) { -TempVal + = L2Val; -L2 = L2Next; -             } +             if(Carry = =1) { -tempval++; +             } A             //without the If Judgment atresult, val = tempval%Ten; -carry = tempval/Ten; -         } -         returnDummy.next; -     } -};

Leetcode2 ADD Numbers