题目：
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.
spoilers alert... click to show 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.
代码：
class Solution {
public:
int myAtoi(string str) {
const size_t len = str.length();
// index of str
size_t i = ;
// skip the white space
while ( str[i]==' ' && i<len ) i++;
int sign = ;
if ( str[i]=='+' ) {
sign = ;
++i;
}
else if( str[i]=='-' ){
sign = -;
++i;
}
// visit all the other char
int num = ;
while ( i < len )
{
// not numerical char
if ( str[i]<'' || str[i]>'' ) break;
// overflow max bound or min bound
if ( num > INT_MAX/ || ( num==INT_MAX/ && (str[i]-'')>INT_MAX%))
{
return sign== ? INT_MAX : INT_MIN;
}
// accumulate the num
num = num * + str[i] - '';
++i;
}
// return the result
return num * sign;
}
};
Tips：
逐步把case处理掉。这里主要有两个点需要关注：
(1)
复习了一下字符串相减，就是其ASCII码相减，这个跟学C语言的时候差不多。
(2)
判断是否超出INT_MAX的时候，一开始写成了“num*10 > INT_MAX”，这种写法是有问题的比如下面的test case就无法通过：
" -11919730356x"
在代码中加一句每次处理num前看看num是什么的语句cout << "num:" << num << ",num*10:" << num*10 <<endl;
num:0,num*10:0
num:1,num*10:10
num:11,num*10:110
num:119,num*10:1190
num:1191,num*10:11910
num:11919,num*10:119190
num:119197,num*10:1191970
num:1191973,num*10:11919730
num:11919730,num*10:119197300
num:119197303,num*10:1191973030
num:1191973035,num*10:-965171538
在这一步就出错了，原因是
2147483647(INT_MAX)
1191973035*10 超出了int类型的限制，就溢出了。
这里有一个问题，为什么溢出的值是-965171538呢？
首先11919730350由十进制转换为二进制是1011 0001100111 1000101001 1010101110 （总共34位数，不包括符号位）
根据题意要求返回的是int型（占4个byte，算第一个符号位在内一共32为）；此时num的最高的两位（34 33）就被舍弃了，最终保留的结果是
num=11 0001100111 1000101001 1010101110
第一位是符号位1（代表负数），其余的31位转换成十进制的数就是96517538.
所以 还是用num > INT_MAX/10这种比较形式安全(保证num不会超过INT_MAX的限制)，以后这种方法也可以沿用。
下面的blog是详细讲C语言整形溢出的。
http://coolshell.cn/articles/11466.html
=============================================
第二次过这道题，最关键的溢出判断的点考虑到了。一些细节，刷了两遍也AC了。
class Solution {
public:
int myAtoi(string str) {
const int len = str.size();
int i = ;
while ( str[i]==' ' && i<len ) ++i;
if ( i==len ) return ;
int sign = ;
if ( str[i]=='+' ){
sign = ;
++i;
}
else if ( str[i]=='-' ){
sign = -;
++i;
}
int num = ;
while ( i<len )
{
if ( str[i]<'' || str[i]>'' ) break;
if ( num>INT_MAX/ || ( num==INT_MAX/ && (str[i]-'')>INT_MAX% ) )
{
return sign== ? INT_MAX : INT_MIN;
}
num = *num + (str[i]-'');
++i;
}
return num * sign;
}
};