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29. Divide Two Integers

题目

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

If it is overflow, return MAX_INT. 

解析

• 此题主要考虑整数的处理，主要考虑正负号，是否溢出
• 另外针对int最小负值取绝对值的用法

class Solution_29 { //此题保证一定除尽的条件
public:
    int divide(int dividend, int divisor) {

        if (divisor == 0||(dividend==INT_MIN&&divisor==-1)) //考虑越界的问题
        {
            return INT_MAX;
        }
        int ret = 0;
        bool sign = (dividend > 0) ^ (divisor > 0); //异号为1

        //long long dividend_ = labs(dividend); //转换 abs(-2147483648)=-2147483648 ;换成labs在leetcode AC过但是VS2013编译器还是负数
        //long long divisor_ = labs(divisor);  //区别lab，abs,fabs()

        //long long dividend_ = llabs(long long(dividend)); //转换 abs(-2147483648)=-2147483648 //强制类型转换一个long和int 一样的，必须有两个long long

                long long dividend_ = llabs((dividend)); //用llabs()即可
        long long divisor_ = llabs(divisor);  //区别lab，abs,fabs()
        while (dividend_>=divisor_)
        {
            int temp = divisor_;
            int multi = 1;//被除数的次数
            while (dividend_ >= (temp << 1)) //成倍的增加，时间更快
            {
                temp=temp << 1;
                multi=multi << 1;
            }
            dividend_ -= temp;
            ret += multi;
        }
        
        return sign?-ret:ret;
    }
};

    Solution_29 su_29;
    su_29.divide( 0-2147483648, 1);

题目来源

• 29. Divide Two Integers