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