8. String to Integer

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.

将一个字符串转化为 int 型。

1. null or empty string
2. white spaces
3. +/- sign
4. calculate real value
5. handle min & max

public int atoi(String str) {
    if (str == null || str.length() < 1)
        return 0;
 
    // trim white spaces
    str = str.trim();
 
    char flag = '+';
 
    // check negative or positive
    int i = 0;
    if (str.charAt(0) == '-') {
        flag = '-';
        i++;
    } else if (str.charAt(0) == '+') {
        i++;
    }
    // use double to store result
    double result = 0;
 
    // calculate value
    while (str.length() > i && str.charAt(i) >= '0' && str.charAt(i) <= '9') {
        result = result * 10 + (str.charAt(i) - '0');
        i++;
    }
 
    if (flag == '-')
        result = -result;
 
    // handle max and min
    if (result > Integer.MAX_VALUE)
        return Integer.MAX_VALUE;
 
    if (result < Integer.MIN_VALUE)
        return Integer.MIN_VALUE;
 
    return (int) result;
}

class Solution:
    # @return an integer
    def atoi(self, str):
        str = str.strip()
        if not str:
            return 0

        MAX_INT = 2147483647
        MIN_INT = -2147483648
        ret = 0
        overflow = False
        pos = 0
        sign = 1

        if str[pos] == '-':
            pos += 1
            sign = -1
        elif str[pos] == '+':
            pos += 1

        for i in range(pos, len(str)):
            if not str[i].isdigit():
                break
            ret = ret * 10 + int(str[i])
            if not MIN_INT <= sign * ret <= MAX_INT:
                overflow = True
                break

        if overflow:
            return MAX_INT if sign == 1 else MIN_INT
        else:
            return sign * ret

# 正则表达式
class Solution:
    # @return an integer
    def atoi(self, str):
        str = str.strip()
        str = re.match(r'^[+-]?d+', str).group()
        MAX_INT = 2147483647
        MIN_INT = -2147483648

        try:
            ret = int(str)
            if ret > MAX_INT:
                return MAX_INT
            elif ret < MIN_INT:
                return MIN_INT
            else:
                return ret
        except:
            return 0