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++

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function signature accepts a

const char *

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spoilers alert... click to show requirements for atoi.

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.
题意：实现atoi，函数应能过滤无效输入（空格为有效输入），且大于int最大范围时输出int最大值，小于int最小值时返回int最小值，异常情况返回零。
思路：比较。利用负数的绝对值比正数的绝对值大1

class Solution {
public:
int myAtoi(string str) {
if (str.empty())
return 0;
if (str[0]!= ' ' && str[0] != '-' && str[0] != '+' && (str[0] - '0' < 0 || str[0] - '0' > 9))
return 0;
int index = 0;
bool negative = false;
while (str[index] == ' ')
index++;
if (str[index] == '-'){
negative = true;
index++;
}
else if (str[index] == '+')
index++;
else{}
long long int n = 0;
while (index < str.size()){
int a = str[index] - '0';
if (a < 0 || a > 9){
if (negative)
n = -n;
return n;
}
n = n * 10 + a;
if (negative == false && n>(numeric_limits<int>::max)())
return (numeric_limits<int>::max)();
if (negative == true && n > (long long int)((numeric_limits<int>::max)()) + 1)
return (numeric_limits<int>::min)();
index++;
}
if (negative)
n = -n;
return (int)n;
}
};