Divide two integers without using multiplication, division and mod operator.
思路：这道题是在别人的提示下用二分搜索才想到的。设被除数为a,除数为b,主要是找出a落在[2^k*b,2^(k+1)*b]的区间，左侧小于a，右侧大于a。然后设置一个指针，为2^(k-1)*b，判断a落在[2^k*b, 2^k*b+2^(k-1)*b]、[2^k*b+2^(k-1)*b,2^(k+1)*b]的哪个区间，然后再设置指针为2^(k-2)*b，继续上述步骤，区间逼近于a。
class Solution {
public:
int toPositive(long long n, bool positive)
{
if (!positive)
{
n = 0 - n;
}
return n;
}
int divide(int dividend, int divisor) {
bool positive = true;
long long longDividend = dividend;
long long longDivisor = divisor;
if (dividend < 0 && divisor < 0)
{
longDividend = 0 - (long long)dividend;
longDivisor = 0 - (long long)divisor;
}
else if (dividend < 0 && divisor > 0)
{
positive = false;
longDividend = 0 - (long long)dividend;
}
else if (dividend >= 0 && divisor < 0)
{
positive = false;
longDivisor = 0 - (long long)divisor;
}
long long i = 1;
long long divid = longDivisor,num;
long long res = 0;
while(divid < longDividend)
{
i = i << 1;
divid = divid << 1;
}
if (divid == longDividend)
{
return toPositive(i, positive);
}
divid = divid >> 1;
i = i >> 1;
num = divid;
res = i;
while(divid > longDivisor)
{
divid = divid >> 1;
i = i >> 1;
if (num + divid <= longDividend)
{
res += i;
num += divid;
}
}
return toPositive(res, positive);
}
};