[LeetCode] 29. Divide Two Integers 两数相除
Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.
求两数相除，不能用乘法，除法和取余操作。
解法：位操作Bit Operation, Time: O(logn)
num=a_0*2^0+a_1*2^1+a_2*2^2+...+a_n*2^n
Java:
public int divide(int dividend, int divisor) {
//handle special cases
if(divisor==0) return Integer.MAX_VALUE;
if(divisor==-1 && dividend == Integer.MIN_VALUE)
return Integer.MAX_VALUE;
//get positive values
long pDividend = Math.abs((long)dividend);
long pDivisor = Math.abs((long)divisor);
int result = 0;
while(pDividend>=pDivisor){
//calculate number of left shifts
int numShift = 0;
while(pDividend>=(pDivisor<<numShift)){
numShift++;
}
//dividend minus the largest shifted divisor
result += 1<<(numShift-1);
pDividend -= (pDivisor<<(numShift-1));
}
if((dividend>0 && divisor>0) || (dividend<0 && divisor<0)){
return result;
}else{
return -result;
}
}
Python:
class Solution:
def divide(self, dividend, divisor):
"""
:type dividend: int
:type divisor: int
:rtype: int
"""
result, dvd, dvs = 0, abs(dividend), abs(divisor)
while dvd >= dvs:
inc = dvs
i = 0
while dvd >= inc:
dvd -= inc
result += 1 << i
inc <<= 1
i += 1
if dividend > 0 and divisor < 0 or dividend < 0 and divisor > 0:
return -result
else:
return result
if __name__ == "__main__":
print(Solution().divide(123, 12))
print(Solution().divide(123, -12))
print(Solution().divide(-123, 12))
print(Solution().divide(-123, -12))
Python:
class Solution:
def divide(self, dividend, divisor):
"""
:type dividend: int
:type divisor: int
:rtype: int
"""
positive = (dividend < 0) is (divisor < 0)
dividend, divisor = abs(dividend), abs(divisor)
res = 0
while dividend >= divisor:
temp, i = divisor, 1
while dividend >= temp:
dividend -= temp
res += i
i <<= 1
temp <<= 1
if not positive:
res = -res
return min(max(-2147483648, res), 2147483647)
C++:
class Solution {
public:
int divide(int dividend, int divisor) {
long long m = abs((long long)dividend), n = abs((long long)divisor), res = 0;
if (m < n) return 0;
while (m >= n) {
long long t = n, p = 1;
while (m > (t << 1)) {
t <<= 1;
p <<= 1;
}
res += p;
m -= t;
}
if ((dividend < 0) ^ (divisor < 0)) res = -res;
return res > INT_MAX ? INT_MAX : res;
}
};
C++:
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)) res = -res;
return res > INT_MAX ? INT_MAX : res;
}
};
