【LeetCode】8. String to Integer (atoi) 字符串转换为整数

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.
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.

理解：

即将一个整数字符串转换为整数，要考虑以下几种情况：

（1）字符串最前面多余的空格，如“ 123”；

（2）数字前面可能带有正号‘+’，或者‘-’，或者不带正负号，如“ +123”，“-123”，“123”；

（3）数字字符串中间可能带有非数字字符，此时返回前面的结果，如“ -000123@123”需要返回-123；

（4）考虑越界的情况，若大于整型的最大值，则返回整型的最大值；若小于整型的最小值，则返回整型的最小值。

代码：

class Solution {
public:
int myAtoi(string str) {
long long integer=0;//为防止越界，返回的整数设置为long long类型

int i=0;
while(i<str.size()&&str[i]==' ')//
i++;

int flag=0;//0表示正数，1表示负数
if(str[i]=='-')
{
flag=0;//为负数
i++;
}
else if(str[i]=='+')
{
flag=1;//为正数
i++;
}
else if(str[i]>='0'&&str[i]<='9')
flag=1;//为正数，此时i已经指向第一个数字，无需执行i++操作

while(str[i]>='0'&&str[i]<='9')
{
integer=integer*10+((int)str[i]-48);
if(integer>INT_MAX&&flag==1)//越界判断，判断是否大于整数最大值
return INT_MAX;
if(-integer<INT_MIN&&flag==0)//越界判断，判断是否小于整数最小值
return INT_MIN;
i++;
}

if(flag==0)
return -integer;
return integer;
}
};