leetcode: Divide Two Integers

http://oj.leetcode.com/problems/divide-two-integers/

Divide two integers without using multiplication, division and mod operator.

思路：

典型的二分法。以87除4举例， (4 * 2 = 8) => (8 * 2 = 16) => (16 * 2 = 32) => (32 * 2) => 64，因为64 * 2 = 128大于87，现在我们可以确定4 * 16 = 64小于87，那么再处理87 - 64 = 23，23除4的话用上面方法可以得到5，还余3，因为小于4，扔掉，所以最后结果是16 + 5 = 21。因为需要考虑溢出问题，所以先把数字转换为unsigned long long，最后转成int返回。

class Solution {
public:
long long internalDivide(unsigned long long dividend, unsigned long long divisor) {
if (dividend < divisor) {
return 0;
}

long long result = 1;
unsigned long long tmp = divisor, left;

while (tmp <= dividend) {
left = dividend - tmp;
tmp <<= 1;

if (tmp > dividend) {
break;
}
else {
result <<= 1;
}
}

return result + internalDivide(left, divisor);
}

int divide(int dividend, int divisor) {
unsigned long long _dividend = abs((long long)dividend),
_divisor = abs((long long)divisor);
bool positive = ((dividend >= 0) && (divisor > 0)) || ((dividend <= 0) && (divisor < 0));

return positive ? internalDivide(_dividend, _divisor) : (-1) * internalDivide(_dividend, _divisor);
}
};