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## Thursday, October 27, 2016

### LeetCode OJ - Sum of Two Integers

Problem:

Please find the problem here.

Analysis:

There are many ways to cheat without using the '+' operator, for example, we could do

$\log(e^x \times e^y) = \log(e^x) + \log(e^y) = x + y$

But instead, I implemented an adder instead.

Solution:

It is relatively easy to implement a ripple carry adder. The implementation is interesting because if I cannot use the '+' operator, I can't even write a loop. So I solved the problem with a simple macro.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/sum-of-two-integers/

#include "LEET_SUM_OF_TWO_INTEGERS.h"
#include <map>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_SUM_OF_TWO_INTEGERS
{
class Solution {
public:
int getSum(int a, int b)
{
int result = 0;
int carry = 0;
int mask = 1;
int bit1;
int bit2;

#define LINE \
bit1 = a & mask;                                                                                                          \
bit2 = b & mask;                                                                                                          \
if (bit1 != 0 && bit2 == 0 && carry == 0 || bit1 == 0 && bit2 != 0 && carry == 0 || bit1 == 0 && bit2 == 0 && carry != 0) \
{                                                                                                                         \
result |= mask; carry = 0;                                                                                            \
}                                                                                                                         \
else if (bit1 != 0 && bit2 != 0 && carry == 0 || bit1 != 0 && bit2 == 0 && carry == bit1 == 0 && bit2 != 0 && carry != 0) \
{                                                                                                                         \
carry = 1;                                                                                                            \
}                                                                                                                         \
else if (bit1 != 0 && bit2 != 0 && carry != 0)                                                                            \
{                                                                                                                         \
result |= mask; carry = 1;                                                                                            \
}                                                                                                                         \
mask = mask << 1;                                                                                                         \

LINE LINE LINE LINE LINE LINE LINE LINE
LINE LINE LINE LINE LINE LINE LINE LINE
LINE LINE LINE LINE LINE LINE LINE LINE
LINE LINE LINE LINE LINE LINE LINE LINE

return result;
}
};
};

using namespace _LEET_SUM_OF_TWO_INTEGERS;

int LEET_SUM_OF_TWO_INTEGERS()
{
Solution solution;
cout << solution.getSum(32032, 2132) << " " << 32032 + 2132 << endl;

return 0;
}