Leetcode OJ #2 Add-Numbers

https://leetcode.com/problems/add-two-numbers/

Test instructions: Two linked lists represent two numbers,

2 4 3

5 6 4

Follow the rules of addition and carry them from left to right.

Output: and the linked list.

is still a water problem. Some special situations need to be considered (I think about it when I do the problem)

1, if two lengths, such as L1>L2, then need to L1 more than L2 out of the part plus carry copy into the ANS

2, 5+5, although the length is equal, but the most high

Unfortunately, WA had two rounds, because

The right one should be

int v = l1->val + L2->val +add;ans, val = v%10;

I write

int v = l1->val + L2->val;ans, val = v%10+add;

Real Two pen ... In addition, the data structure is really not ripe, you have to write an analysis of STL source code ...

Here is the test program

#include <cstdio> #include <cstring> #include <vector> #include <algorithm> #include < Iostream>using namespace Std;      struct ListNode {int val;      ListNode *next; ListNode (int x): Val (x), Next (NULL) {}}; ListNode * a=null, * b=null;class Solution {public:listnode* addtwonumbers (listnode* L1, listnode* L2) {int ad        d = 0;        listnode* ans = 0;        listnode* ret=0;        int f = 0; if (l1| |        L2) ans = new ListNode (0);            while (L1 && L2) {int v = l1->val + l2->val +add;            cout << "DEBUG" << l1->val << "<< l2->val << Endl;                 if (f = = 0) {ret = ans;                 f++;            ans, val = v%10;                }else{ans-next = new ListNode (0);                Ans = ans->next;                ans, val = v%10;            Ans->next = 0;            } add = V/10; L1 =l1->next;        L2 = l2->next;            }//finally need to process add if (L1 | | L2) {listnode* remain = L1? l1:l2;                while (remain) {//cout << "DEBUG" << remain->val << Endl;                int v = remain->val+add;//+ l2->val;                Ans->next = new ListNode (V%10);                add = V/10;                Ans = ans->next;                Ans->next = NULL;            remain = remain->next;                }} if (add) {ans->next = new ListNode (add);            Ans->next->next = 0;    } return ret;    }};int Main () {//freopen ("In.txt", "R", stdin);    String str1, str2;    CIN >> str1 >> str2;    ListNode *sta=null, *stb=null;    if (Str1.size ()) sta = a = new ListNode (0);        for (int i=0;i<str1.size (); i++) {a->val = str1[i]-' 0 '; A->next = (I==str1.size ()-1)?        Null:new ListNode (0);  A = a->next;  } if (Str2.size ()) STB = b = new ListNode (0);        for (int i=0;i<str2.size (); i++) {b->val = str2[i]-' 0 '; B->next = (i = = Str2.size ()-1)?        Null:new ListNode (0);    b = b->next;    } listnode* ans = ((new solution ())->addtwonumbers (STA,STB));//listnode* ans = NULL;        while (ans) {cout << ans->val << Endl;    ans = ans next; } return 0;}

Leetcode OJ #2 Add-Numbers