[LeetCode]String to Integer (atoi)

将string或const char *c转成int，首先找第一个非空字符，如果第一个非空字符是“+-0123456789”说明为有效int，否则输出0

找到非空字符后，如果第一个非空字符为符号位，记录符号，否则，继续。

符号位后的字符不是”0123456789“也是无效的int，输出0

其他情况正确处理

/*
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:
//stoi,stod，stof(string)
int myAtoi(string str) {
string::size_type sz;
for(sz=0; sz!=str.size(); ++sz)     //找非空字符
if(str[sz]==' ')
continue;
else
break;
int flag = 1;   //1 for plus
if(str[sz]=='-'){
flag = -1;
++sz;
}
else if(str[sz]=='+'){
flag = 1;
++sz;
}
char c = str[sz];
long long int ret=0;
if(c<'0'||c>'9')    return 0;   //invalid
while(c>='0'&&c<='9'){
if(flag ==1)
ret = 10*ret + c-'0';
else
ret = 10*ret-(c-'0');
c = str[++sz];
if(ret>INT_MAX)
ret = INT_MAX;
else if(ret<INT_MIN)
ret = INT_MIN;
}
return ret;

}
};