Add the two numbers and return it as a linked list.

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.

	struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
	};
	class Solution {
	unsigned long long getListSum(ListNode* l)
	{
	ListNode *trav=l;
	unsigned long long sum=0;
	unsigned long long powPos=1;
	while(NULL!=trav)
	{
	sum=sum+((trav->val)*powPos);
	powPos*=10;
	trav=trav->next;
	}
	return sum;
	}
	void InsertAtBack(ListNode **res,int item)
	{
	ListNode* newNode=new ListNode(item);
	ListNode* last=*res;
	if(nullptr==*res)
	{
	*res=newNode;
	}
	else
	{
	while (last->next != nullptr)
	last = last->next;
	last->next = newNode;
	}
	}
	public:
	ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
	ListNode* result;
	ListNode *l1Itr=l1;
	ListNode *l2Itr=l2;
	ListNode** itr=&result;
	int num=0;
	unsigned int carryForwardSum=0;
	while(nullptr!=l1Itr || nullptr!=l2Itr)
	{
	if(nullptr!=l1Itr)
	{
	num+=l1Itr->val;
	l1Itr=l1Itr->next;
	}
	if(nullptr!=l2Itr)
	{
	num+=l2Itr->val;
	l2Itr=l2Itr->next;
	}
	num+=carryForwardSum;
	if(num>=10)
	{
	carryForwardSum=num/10;
	}
	else
	{
	carryForwardSum=0;
	}
	num=num%10;
	(*itr)=new ListNode(num);
	itr=&((*itr)->next);
	num=0;
	}
	if(carryForwardSum!=0 && l1Itr==nullptr && l2Itr==nullptr)
	{
	(*itr)=new ListNode(carryForwardSum);
	itr=&((*itr)->next);
	carryForwardSum=0;
	}
	return result;
	}
	};

view raw AddTwoNumbers_InList.cpp hosted with ❤ by GitHub

1. Two Sum

Problem:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,

return [0, 1].

Solution:

class Solution {

public:

    vector<int> twoSum(vector<int>& nums, int target) {

        vector<int> res(2);

        int size=nums.size();

        std::map<int,int> hashMap;

        for(int i=0;i<size;i++){

            int curr=target-nums[i];

            if(hashMap.find(curr)!=hashMap.end())

            {

                res[0]=hashMap[curr];

                res[1]=i;return res;

            }

            else

            {

                hashMap[nums[i]]=i;

            }

        }

        return res;     

    }

};

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