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

If it is overflow, return 2147483647

Have you met this question in a real interview?

 
 
Example

Given dividend = 100 and divisor = 9, return 11.

LeetCode上的原题,请参见我之前的博客Divide Two Integers

解法一:

class Solution {
public:
/**
* @param dividend the dividend
* @param divisor the divisor
* @return the result
*/
int divide(int dividend, int divisor) {
if (divisor == || (dividend == INT_MIN && divisor == -)) return INT_MAX;
long long m = abs((long long)dividend), n = abs((long long)divisor), res = ;
int sign = ((dividend < ) ^ (divisor < )) ? - : ;
if (n == ) return sign == ? m : -m;
while (m >= n) {
long long t = n, p = ;
while (m >= (t << )) {
t <<= ;
p <<= ;
}
res += p;
m -= t;
}
return sign == ? res : -res;
}
};

解法二:

class Solution {
public:
/**
* @param dividend the dividend
* @param divisor the divisor
* @return the result
*/
int divide(int dividend, int divisor) {
long long m = abs((long long)dividend), n = abs((long long)divisor), res = ;
if (m < n) return ;
while (m >= n) {
long long t = n, p = ;
while (m > (t << )) {
t <<= ;
p <<= ;
}
res += p;
m -= t;
}
if ((dividend < ) ^ (divisor < )) res = -res;
return res > INT_MAX ? INT_MAX : res;
}
};

解法三:

class Solution {
public:
/**
* @param dividend the dividend
* @param divisor the divisor
* @return the result
*/
int divide(int dividend, int divisor) {
long long m = abs((long long)dividend), n = abs((long long)divisor), res = ;
if (m < n) return ;
long long t = n, p = ;
while (m > (t << )) {
t <<= ;
p <<= ;
}
res += p + divide(m - t, n);
if ((dividend < ) ^ (divisor < )) res = -res;
return res > INT_MAX ? INT_MAX : res;
}
};

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