DivideTwoIntegers.cpp

class Solution {
public:
    int divide(int dividend, int divisor) {
      //First check whether the final result will encounter overflow
        if(!divisor || (dividend == INT_MIN && divisor == -1))
            return INT_MAX;
      //We use long to store the final reslut, 
      //because we convert the negative number to positive
      //If dividend is -2^31, we need to make sure that we will not have overflow
        long divd = labs(dividend);
        long divs = labs(divisor);
      //Save the sign, if both of them or neither of them are negative,
      //bit xor operation will return flase, sign will be 1.
        int sign = ((dividend<0)^(divisor<0))?-1:1;
        int finalRes = 0;
        while(divd>=divs){
            long temp = divs; 
            int multiple = 1;
          //We update temp to 2* temp every time, in oder to save time
          //We use the myltiple to keep track of how many times we have shifted our temp,
          //We use bit left shift to synchrosize the operation, for each for loop, 
          //we can say that 2 * previous multiple divisor can be subtracted from the dividend.
          //multiple will add to final result
            while(divd>=(temp<<1)){
                temp = temp<<1;
                multiple = multiple<<1;
            }
          //If dividend - temp is still greater than divisor, we need to calculate from multiple = 1 again.
            divd -= temp;
            finalRes += multiple;            
        }
        return sign>0? finalRes:-finalRes;  
    }
};