今天的两道题关于基本数据类型的探讨，估计也是要考虑各种情况，要细致学习
7. Reverse Integer
Reverse digits of an integer.
Example1: x = 123, return 321 Example2: x = -123, return -321
Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
If the integer‘s last digit is 0, what should the output be? ie, cases such as 10, 100.
Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?
For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
Update (2014-11-10): Test cases had been added to test the overflow behavior.
注：颠倒整数中的每一位，符号位保留。估计会有一些极端情况需要考虑，这里指出了几个，如，末位为0 翻转之后是什么？又如，如果翻转之后超出了数据类型的表达范围怎么办？。。。可能还会遇到其它的困难^~^
解：在考虑最值的时候必须另外定义最值为int 常量，否则的话系统会给他分配一个合适的类型
直接写成这个做判断是不可以的。
if (temp >0x7fffffff || temp < 0x80000000)
{
return result;
}
下面是提交通过的代码：
class Solution {
public:
int reverse(int x) {
int result{0};
long long temp{0};
const int max_int=0x7fffffff;// 0111 1111 1111 1111 ... 1111 32bit
const int min_int=0x80000000;//1000 0000 0000 0000 ... 0000 32bit
while (x != 0)
{
temp = temp*10 + x%10;
x = x / 10;
if (temp > max_int || temp < min_int)
{
return result;
}
}
result = temp;
return result;
}
};
8. String to Integer (atoi)
Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
Update (2015-02-10): The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.
Requirements for atoi:
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.
The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.
If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.
If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.
注：把一个字符串转换为整数。要考虑所有可能的输入情况。这个问题在《c++入门经典》中设计计算器时遇到过，只是当时没有细细的分析每一种情况，现在要好好看看。
解：这个简直是不可理喻，一下列举了一些可能出现的情况
input：“+ - 2”expected：0
input：“+ 01 1 2”expected：1
input：“+ + 2”expected：0
input：“-012a34”expected：12
input：" +11191657170"expected：2147483647
代码如下，有待精简：
long long result{ 0 };
int n{ 0 };//位数
int sign{ 1 }, p_n_flag{ 0 }, zero_flag{0};
for (string::iterator s_i = str.begin(); s_i != str.end(); s_i++)
{
if (*s_i == ‘-‘)
{
if (p_n_flag==0)
{
p_n_flag = 1;
sign = -1;
continue;
}
else
{
sign = 0;//如果出现两次被认为是非法输入
break;
}
}
if (*s_i == ‘+‘)
{
if (p_n_flag == 0)
{
p_n_flag = 1;
sign = 1;
continue;
}
else
{
sign = 0;
break;
}
}
if (*s_i == ‘ ‘)
{
if (n == 0 && zero_flag == 0 && p_n_flag==0)
{ continue; }
else
{break;}
}
if (*s_i == ‘0‘)
{
zero_flag = 1;
if (n == 0)
{continue;}
else
{
n = n + 1;
result = result * 10;
if (result > INT_MAX)
{
switch (sign)
{
case 0:return 0;
case 1:return INT_MAX;
case -1:return INT_MIN;
default:
break;
}
}
else
{continue;}
}
}
if ((*s_i) > ‘0‘ && ((*s_i) < ‘9‘ || (*s_i) == ‘9‘))
{
n = n + 1;
//result = result * 10 + ((*s_i) - 48);
result = result * 10 + ((*s_i) - ‘0‘);//最后一个括号里不用显式的值48 而是用‘0’
cout << result << endl;
if (result > INT_MAX)
{
switch (sign)
{
case 0:return 0;
case 1:return INT_MAX;
case -1:return INT_MIN;
default:
break;
}
}
}
else
{
break;
}
}
result = result*sign;
return result;