Leetcode 2-add two Numbers (c + + Java Python)

Title: http://oj.leetcode.com/problems/add-two-numbers/

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

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

Title translation:

A given two lists represent two non-negative numbers. The numbers are stored in reverse order, and each node contains a single number. Calculates the two-digit number and returns it in the form of a linked list. Analysis:

To consider rounding. The node that represents the result is to be new.

C + + implementation:

/** * 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) {//Important:please reset A

		NY member data you declared, AS//the same Solution instance'll be reused for each test case.
		if (L1 = = NULL) {return L2;
		} if (L2 = = NULL) {return L1;
		} ListNode *result = NULL;

		ListNode *sum = NULL;
		int val = 0;

		int carry = 0;
			while (L1!= null | | | L2!= NULL) {val = carry;
			if (L1!= NULL) {val + = l1->val;
			} if (L2!= NULL) {val + = l2->val;
			} carry = VAL/10;

			Val-= carry * 10;
				if (sum = = NULL) {sum = new ListNode (val);
			result = SUM;
				else {sum->next = new ListNode (val);
			sum = sum->next;
			} if (L1!= NULL) {L1 = l1->next; } if (L2!= NULL) {L2 = L2->next;
		} if (carry!= 0) {sum->next = new ListNode (carry);
    return result; }
};

Java implementation: (unlike C + + implementations)

/** * Definition for singly-linked list.
 * public class ListNode {* int val;
 * ListNode Next;
 * ListNode (int x) {* val = x;
 * next = NULL; *} */public class Solution {public ListNode addtwonumbers (listnode L1, ListNode L2) {//Importan
		T:please Reset any member of the data you declared, AS//the same Solution instance'll be reused to each test case.
		if (L1 = = null) {return L2;
		} if (L2 = = null) {return L1;
		int len1 = 0;

		int len2 = 0;

		ListNode head = L1;
			while (head!= null) {++len1;
		head = Head.next;

		head = L2;
			while (head!= null) {++len2;
		head = Head.next; } listnode longer = Len1 >= len2?
		L1:L2; ListNode shorter = Len1 < len2?

		L1:L2;
		ListNode result = null;

		ListNode sum = null;
		int val = 0;

		int carry = 0;
			while (shorter!= null) {val = longer.val + shorter.val + carry;
			carry = VAL/10;
			
			Val-= carry * 10; if (sum == null) {sum = new ListNode (val);
			result = SUM;
				else {sum.next = new ListNode (val);
			sum = Sum.next;
			} longer = Longer.next;
		shorter = Shorter.next;
			while (longer!= null) {val = longer.val + carry;
			carry = VAL/10;
			
			Val-= carry * 10;
			Sum.next = new ListNode (val);
			
			sum = Sum.next;
		longer = Longer.next;
		} if (carry!= 0) {sum.next = new ListNode (carry);
    return result; }
}

Python implementation:

# Definition for singly-linked list. # class ListNode: # def __init__ (self, x): # self.val = x # self.next = None class Solution: # @r
            Eturn a ListNode def addtwonumbers (self, L1, L2): If L1 = None:return L2 If L2 = None:
            return L1 len1 = 0 Len2 = 0 head = L1 while head!= None: Len1 + 1 head = Head.next head = L2 while head!= None:len2 +
        = 1 head = Head.next if len1 >= len2:longer = L1 shorter = L2
            Else:longer = L2; shorter = L1 sum = None carry = 0 while shorter!= none:va
            
            Lue = Longer.val + Shorter.val + carry carry = VALUE/10 value-= carry * 10
  if sum = = None:sum = ListNode (value)              result = Sum Else:sum.next = ListNode (value) sum = Sum.next longer = longer.next shorter = Shorter.next while longer!= Non
            
            E:value = Longer.val + carry carry = VALUE/10 value-= carry * 10
            
        Sum.next = ListNode (value) sum = sum.next longer = Longer.next If carry!= 0:sum.next = ListNode (carry) return result

        Thank you for reading and welcome comments.