leetcode 8. String to Integer (atoi)

Implement atoi to
convert a string to an integer.

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.
将字符串转化成整形,去掉前导空格和0
如果结果溢出,则返回int的最大数(最小数)

如果不能转换,则返回0

public class A8StringtoInteger_atoi {

public int myAtoi(String str) {
if(str == null || "".equals(str = str.trim()))
return 0;
int sign = 1, i = 0;
long ans = 0L;
if(str.charAt(0) == '-') {
i++;
sign = -1;
} else if (str.charAt(0) == '+') {
i++;
}
for(; i < str.length(); i++) {
if(str.charAt(i) >= '0' && str.charAt(i) <= '9') {
ans = ans * 10L + (str.charAt(i) - '0');
if(ans * sign >= (long)Integer.MAX_VALUE)
return Integer.MAX_VALUE;
if(ans * sign <= (long)Integer.MIN_VALUE)
return Integer.MIN_VALUE;
} else
break;
}
return (int)ans * sign;
}
}