Given two integers dividend and divisor, divide two integers without using multiplication, division and mod operator.
Return the quotient after dividing dividend by divisor.
The integer division should truncate toward zero.
Example 1:
Input: dividend = 10, divisor = 3 Output: 3
Example 2:
Input: dividend = 7, divisor = -3 Output: -2
Note:
- Both dividend and divisor will be 32-bit signed integers.
- The divisor will never be 0.
- Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 231 − 1 when the division result overflows.
考察：int類型兩數整除結果，轉化爲高階long long類型來運算，如果溢出則返回INT_MAX, 儘可能去使用位移動，以及異或運算求符號；
class Solution {
public:
int divide(int dividend, int divisor) {
long long res = 0;
long long m = abs((long long)dividend), n = abs((long long)divisor);
if (m < n)
return 0;
long long t = n, p = 1;
while (m > (t << 1)) {
t <<= 1;
p <<= 1;
}
res += p + divide(m-t, n);
if ((dividend < 0) ^ (divisor < 0)) //符號取異或
res = -res;
return res > INT_MAX ? INT_MAX : res; // 判斷是否溢出
}
};
完，