LeetCode 2 Add Two Numbers

LeetCode 2 Add Two Numbers

Translation:

Two linked lists that represent two non-negative numbers. Numbers are stored in reverse order, and their nodes contain a single number. Add the two numbers and return them as a linked list. Input: (2-> 4-> 3) + (5-> 6-> 4) Output: 7-> 0-> 8

Original question:

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

C ++

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {          int carry=0;        ListNode* listNode=new ListNode(0);        ListNode* p1=l1,*p2=l2,*p3=listNode;        while(p1!=NULL|p2!=NULL)        {            if(p1!=NULL)            {                carry+=p1->val;                p1=p1->next;            }            if(p2!=NULL)            {                carry+=p2->val;                p2=p2->next;            }            p3->next=new ListNode(carry%10);            p3=p3->next;            carry/=10;        }        if(carry==1)            p3->next=new ListNode(1);        return listNode->next;      }};

C #

/** * Definition for singly-linked list. * public class ListNode { *     public int val; *     public ListNode next; *     public ListNode(int x) { val = x; } * } */public class Solution{    public ListNode AddTwoNumbers(ListNode l1, ListNode l2)    {        int carry = 0;        ListNode listNode= new ListNode(0);        ListNode p1 = l1, p2 = l2, p3 = listNode;        while (p1 != null || p2 != null)        {            if (p1 != null)            {                carry += p1.val;                p1 = p1.next;            }            if (p2 != null)            {                carry += p2.val;                p2 = p2.next;            }            p3.next = new ListNode(carry % 10);            p3 = p3.next;            carry /= 10;        }        if (carry == 1)            p3.next = new ListNode(1);        return listNode.next;    }}

Java

public class Solution {     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {         int carry=0;         ListNode listNode=new ListNode(0);         ListNode p1=l1,p2=l2,p3=listNode;         while(p1!=null||p2!=null){             if(p1!=null){                 carry+=p1.val;                 p1=p1.next;             }             if(p2!=null){                 carry+=p2.val;                 p2=p2.next;             }             p3.next=new ListNode(carry%10);             p3=p3.next;             carry/=10;         }         if(carry==1)             p3.next=new ListNode(1);         return listNode.next;       }}

C ++ (from network)

class Solution {public:    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {        int carry=0;        ListNode* res=new ListNode(0);        ListNode* head = res;        while (l1 && l2){            res->next=new ListNode((l1->val+l2->val+carry)%10);            carry = (l1->val+l2->val+carry)/10;            l1=l1->next;            l2=l2->next;            res=res->next;        }                while (l1){            res->next=new ListNode((l1->val+carry)%10);            carry = (l1->val+carry)/10;            l1=l1->next;            res=res->next;        }                while (l2){            res->next=new ListNode((l2->val+carry)%10);            carry = (l2->val+carry)/10;            l2=l2->next;            res=res->next;        }                if (carry>0){            res->next = new ListNode(carry);        }                return head->next;            }};