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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.
class Solution {
public:
// 字符串转int，注意判断超出Int范围
int myAtoi(string str) {
if (str == "") return 0;
// 去首尾空格
str.erase(0, str.find_first_not_of(‘ ‘));
str.erase(str.find_last_not_of(‘ ‘) + 1);
int i = 0, len = str.length(), sign = 1;
while ( i < len && str[i] == ‘ ‘) i++;
if (i == len) return 0;
if (str[i] == ‘+‘) {
sign = 1;
i++;
} else
if (str[i] == ‘-‘) {
sign = -1;
i++;
}
// 转换结果可能超出int范围，使用long long作为转换后接收变量的类型
long long ret = 0;
for (; i < len; i++) {
if (str[i] < ‘0‘ || str[i] > ‘9‘) break;
ret = ret * 10 + (str[i] - ‘0‘);
if (ret > INT_MAX) break;
}
ret *= sign;
if (ret > INT_MAX) return INT_MAX;
if (ret < INT_MIN) return INT_MIN;
return ret;
}
};
注释：
erase函数的原型如下：
（1）string& erase ( size_t pos = 0, size_t n = npos );
（2）iterator erase ( iterator position );
（3）iterator erase ( iterator first, iterator last );
也就是说有三种用法：
（1）erase(pos,n); 删除从pos开始的n个字符，比如erase(0,1)就是删除第一个字符
（2）erase(position);删除position处的一个字符(position是个string类型的迭代器)
（3）erase(first,last);删除从first到last之间的字符（first和last都是迭代器）
以下是库函数的定义：
stl:
template<typename _Category, typename _Tp, typename _Distance = ptrdiff_t,
typename _Pointer = _Tp*, typename _Reference = _Tp&>
struct iterator
{
/// One of the @link iterator_tags tag types@endlink.
typedef _Category iterator_category;
/// The type "pointed to" by the iterator.
typedef _Tp value_type;
/// Distance between iterators is represented as this type.
typedef _Distance difference_type;
/// This type represents a pointer-to-value_type.
typedef _Pointer pointer;
/// This type represents a reference-to-value_type.
typedef _Reference reference;
};
string:
iterator
erase(iterator __first, iterator __last);
#if __cplusplus >= 201103L
/**
* @brief Remove the last character.
*
* The string must be non-empty.
*/
void
pop_back() // FIXME C++11: should be noexcept.
{ erase(size()-1, 1); }
erase(iterator __first, iterator __last);
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原文地址：http://www.cnblogs.com/wqkant/p/5285916.html
