[LeetCode]29. Divide Two Integers

题目：Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.
题意：求两个数的商；
有两种情况溢出:
1.divisor = 0;
2.dividend = INT_MIN and divisor = -1 (因为 abs(INT_MIN) = INT_MAX + 1).
思路：基本原理是移位，第1次移位的大小正好是divisor的大小，
第2次的大小正好是divisor扩大2倍的大小，第3次的大小正好是divisor扩大2²倍的大小，

第4次的大小正好是divisor扩大2³倍的大小......以此类推；

class Solution {
public:
	int divide(int dividend, int divisor) {
		if (!divisor || (dividend == INT_MIN && divisor == -1))
			return INT_MAX;
		int sign = ((dividend < 0) ^ (divisor < 0)) ? -1 : 1;
		long long dvd = labs(dividend);
		long long dvs = labs(divisor);
		int res = 0;
		while (dvd >= dvs) {
			long long temp = dvs, multiple = 1;
			while (dvd >= (temp << 1)) {
				temp <<= 1;
				multiple <<= 1;
			}
			dvd -= temp;
			res += multiple;
		}
		return sign == 1 ? res : -res;
	}
};