LeetCode Ex08 String to Integer (atoi)

String to Integer (atoi)

Implement atoi which converts 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.

Note:

Only the space character ' ' is considered as whitespace character.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.

Example 1:
Input: "42"
Output: 42

Example 2:
Input: " -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
Then take as many numerical digits as possible, which gets 42.

Example 3:
Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example 4:
Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical digit or a +/- sign. Therefore no valid conversion could be performed.

Example 5:
Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer. Thefore INT_MIN (−231) is returned.

题目

该题要求将字符串转变成整形数，用一种特殊的方法。主要是找到字符串的第一个非空格字符，若是“+或“-”或数字则继续往下找，直到该字符不是数字，最后将其转变成整形返回。若第一个非空字符不满足上面情况，则返回0。需要注意的是，结果可能会溢出。若溢出且该值为正，则返回2^31 − 1；若为负，返回-2^32。

首先我们用str.trim()去除字符串的前后空格，并创建firstChar用来记录第一个字符，若为‘+，sign=1，start++；若为'-'，sign=-1，start++。sign用于修改最后返回时结果的正负，start标记遍历字符串的开始位置。从start开始遍历字符串，若字符不是数字，则返回sum。其中sum=sum*10+str.charAt(i)-'0'，用于记录数字，初始化为0。若sum溢出，则根据上面的情况返回不同的值。

    public int myAtoi(String str) {
        if (str == null || str.length() == 0)
            return 0;//
        str = str.trim();
        if (str.equals(""))
            return 0;
        char firstChar = str.charAt(0);
        int sign = 1, start = 0, len = str.length();
        long sum = 0;
        if (firstChar == '+') {
            sign = 1;
            start++;
        } else if (firstChar == '-') {
            sign = -1;
            start++;
        }
        for (int i = start; i < len; i++) {
            if (!Character.isDigit(str.charAt(i)))
                return (int) sum * sign;
            sum = sum * 10 + str.charAt(i) - '0';
            if (sign == 1 && sum > Integer.MAX_VALUE)
                return Integer.MAX_VALUE;
            if (sign == -1 && (-1) * sum < Integer.MIN_VALUE)
                return Integer.MIN_VALUE;
        }
        return (int) sum * sign;
    }