LeetCode 29. Divide Two Integers

In order to solve this problem, it is good to know some basics first.

Suppose, dividend is a1, divisor is a0.   

a1 = a0 * (2^k) + a0 *(2^(k-1)) + a0*(2^(k-2)) + ... a0 *(2 ^ 1)+ a0*(2^0) +a2...(a2 < a0) if it is dividable, the result is k + k-1 +...0

Several examples can better illustrate this problem.

(9, 4)  9 = (8, 4) + (1, 4), (8, 4) == 2, (1, 4) == 0, Thus, (9, 4) = 2

(15, 4) = (8, 4) + (4, 4) + (3, 4) ---> 3

To find the (8, 4), we just need to shift 4 for k left steps, until it is larger than the dividend. k is the result.

#include <iostream>
#include <climits>
using namespace std;

/*
  Divide two integers without using multiplication, division and mod operator.
  If it is overflow, return MAX_INT.
*/

int divide(int dividend, int divisor) {
  bool negativeFlag = false;
  if(dividend < 0 && divisor >= 0) negativeFlag = true;
  if(dividend >= 0 && divisor < 0) negativeFlag = true;
  int newDividend = abs(dividend);
  int newDivisor = abs(divisor);
  if(newDivisor == 1) return negativeFlag ? -1* newDividend : newDividend;
  if(newDivisor == 0) return INT_MAX;
  if(newDivisor > newDividend || newDividend == 0) return 0;

  int count = 0;
  while(newDividend > 0) {
    int tmp = newDivisor;
    int k = 0;
    while(tmp <= newDividend) {
      k++;
      tmp = tmp << 1;
    }
    count += k;
    newDividend = newDividend - (tmp >> 1);
  }
  return negativeFlag ? -1 * count : count;
}
int main(void) {
  cout << "9 divide 3" << endl;
  cout << "res: " << divide(9, 3) << endl;

  cout << "9 divide 4" << endl;
  cout << "res: " << divide(9, 4) << endl;

  cout << "9 divide -3" << endl;
  cout << "res: " << divide(9, -3) << endl;

  cout << "9 divide 10" << endl;
  cout << "res: " << divide(9, 10) << endl;

  cout << "9 divide 0" << endl;
  cout << "res: " << divide(9, 0) << endl;

  cout << "9 divide 1" << endl;
  cout << "res: " << divide(9, 1) << endl;
}