[leetcode]Divide Two Integers

题目：

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

If it is overflow, return MAX_INT.

描述：实现除法，要求不能使用乘法、除法、取模，注意可能存在溢出

solution by python:

写的略乱。。。

maxint = 2147483647minint = -2147483648class Solution:    # @return an integer    def divide(self, dividend, divisor):        if divisor == 0: return maxint        if divisor == 1: return dividend        if divisor == -1: return -dividend if dividend != minint else maxint        if dividend>0 and divisor<0 or dividend<0 and divisor>0:            ans = self.divide(abs(dividend), abs(divisor))            return -ans if ans!=minint else maxint        dividend = abs(dividend);divisor=abs(divisor)        ans = 0        while dividend >= divisor:            b = 1;d=divisor            while d < dividend>>1:                d <<= 1                b <<= 1            ans += b            dividend -= d        return ans

solution by c++:

class Solution {public:    int divide(int dividend, int divisor) {        bool negative = false;        if(dividend>0 && divisor<0 || dividend<0 && divisor>0) negative = true;        long long dvd = dividend;        long long dvs = divisor;        long long ans = absDivide( abs(dvd), abs(dvs));        if(negative) return 0-ans;        return ans;    }        long long absDivide(long long dividend, long long divisor)    {        if(dividend < divisor) return 0;        long long ans = 1;        long long a = divisor;        while(a <= dividend)        {            ans <<= 1;            a <<= 1;        }        ans >>= 1;        a >>= 1;        return ans+absDivide(dividend-a, divisor);    }};