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 > newDivisor) {   // first round made an error here.    int tmp = newDivisor;    int k = 0;    while(tmp <= newDividend) {      k++;      tmp = tmp << 1;    }    count += 1 << (k-1);                       // made an error here too.    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;}// A recursion version#include <iostream>using namespace std;int divideTwo(long long int divident, long long int divisor) {  if(divident == 0 || divisor > divident) return 0;  if(divisor == 1) return divident;  long long tmp = divisor, k = 0;  for(; divident >= tmp; ++k) {    if(divident - tmp < divisor) {      return 1 << k;    }    tmp <<= 1;  }  return divideTwo(divident - (tmp >> 1), divisor) + (1 << (k - 1));}int divideTwo(int divident, int divisor) {  long long int d0 = divident;  long long int d1 = divisor;  bool negative;  if((d0 > 0 && d1 > 0) || (d0 < 0 && d1 < 0)) negative = false;  else negative = true;  int res = divideTwo(d0, d1);  return negative ? -res : res;}