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/**

* 要求如下：

* 1. 过滤掉前面的空格

* 2. 考虑正负号

* 3. 如果溢出则返回上限或下限

* 4. 解析时遇到非法字符则停止，返回当前结果

* 5，防御空串

*/

/**

* TestCase: ""

*/

public class String2Integer {

public int myAtoi(String str) {

int i = 0, sign = 1;

for ( ; i < str.length() && str.charAt(i) == ' '; i++);

if (i < str.length()) {

char csign = str.charAt(i);

if (csign == '-') {

sign = -1;

i++;

} else if (csign == '+') {

sign = 1;

i++;

} else if (csign < '0' || csign > '9') {

return 0;

} else {}

}

long number = 0;

for ( ; i < str.length(); i++) {

char c = str.charAt(i);

if (c < '0' || c > '9') {

break;

}

number = number * 10 + (c - '0');

if (number * sign > Integer.MAX_VALUE) {

return Integer.MAX_VALUE;

}

if (number * sign < Integer.MIN_VALUE) {

return Integer.MIN_VALUE;

}

}

return (int) (number * sign);

}

}