By yusijia

Updated:

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/*
 * 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.

 * 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.

 */
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public class StringToInteger {

    public int atoi(String str) {
        long result = 0;
        int sign = 1;//1为正-1为负
        int index = 0;
        //找到第一个不是空格的字符
        for(; index < str.length() && str.charAt(index) == ' '; index++) ;
        
        if(index < str.length() && (str.charAt(index) == '+' || str.charAt(index) == '-')){
            sign = str.charAt(index) == '+' ? 1 : -1;
            index++;
        }
        for(; index < str.length(); index++){
            if(str.charAt(index) < '0' || str.charAt(index) > '9')
                break;
            result = result * 10 + (str.charAt(index) - '0');
        }
        result *= sign;
        result = Math.min(result, Integer.MAX_VALUE);
        result = Math.max(result, Integer.MIN_VALUE);
        return (int)result;
    }
}


//题目后来更新了，有的数据超过了long的范围
public int atoi(String str) {
        
    long result = 0;
    int sign = 1;//1为正-1为负
    int index = 0;
    //找到第一个不是空格的字符
    for(; index < str.length() && str.charAt(index) == ' '; index++) ;
    
    if(index < str.length() && (str.charAt(index) == '+' || str.charAt(index) == '-')){
        sign = str.charAt(index) == '+' ? 1 : -1;
        index++;
    }
    //在这里判断时忽略前面的'-'或'+'
    for(; index < str.length(); index++){
        if(str.charAt(index) < '0' || str.charAt(index) > '9')
            break;
        result = result * 10 + (str.charAt(index) - '0');
        //处理大于Integer.MAX_VALUE或小于Integer.MIN_VALUE的数据
        if(result > Integer.MAX_VALUE)  
            return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
    }
    result *= sign;
    result = Math.min(result, Integer.MAX_VALUE);
    result = Math.max(result, Integer.MIN_VALUE);
    return (int)result;
}
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