LeetCode:String to Integer (atoi)

问题描述：

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.

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.

Subscribe to see which companies asked this question

思路：
开始试验了很多次，都是超时，最后发现是对题意不理解，题目的大意是：忽略字符串前面的空格，从第一个非空格字符开始，第一位通常是符号位，首先进行判断，从第二位开始计算，一直到最后一位为止。

代码：

class Solution {
public:
int myAtoi(string str) {
if(str[0] == '\0')
return 0;
int i = 0;
while(str[i] == ' ')
i++;
int flag = 1;
if(str[i] == '+')
{
i++;
}
else if(str[i] == '-')
{
flag = -1;
++i;
}
long  long number = 0;
while(str[i] != '\0')
{
if(str[i] >= '0' && str[i] <= '9')
number = number * 10 + flag * (str[i++] - '0');
else
return number;
if(number > 2147483647 || number < -2147483648)
return number > 0 ? 2147483647 : -2147483648;
}
return (int)number;
}
};