文章目錄

• 1. 題目描述：
• 2. 思路分析：
• 3. Java程式碼：

1. 題目描述：

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: [−2^31, 2^31 − 1]. If the numerical value is out of the range of representable values, INT_MAX (2^31 − 1) or INT_MIN (−2^31) 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 (−2^31) is returned.

2. 思路分析：

題目的意思很簡單，就是將一個字串轉成整型數字。需要注意的是符號問題還有字串頭部空白問題。
思路就是依次解析每個字串，然後比較與字元'0'的差值，即該字元表示的數值大小。
注意： 需要判斷轉換後的數是否越界。

3. Java程式碼：

原始碼：見我GiHub主頁

程式碼：

public static int myAtoi(String str) {
    str = str.trim();

    if (str == null || str.length() < 1) {
        return 0;
    }

    int i = 0;

    // 判斷正負號
    char flag = '+';
    if (str.charAt(0) == '-') {
        flag = '-';
        i++;
    } else if (str.charAt(0) == '+') {
        i++;
    }

    // 用double儲存防止越界
    double result = 0;

    while (i < str.length() && str.charAt(i) >= '0' && str.charAt(i) <= '9') {
        result = result * 10 + (str.charAt(i) - '0');
        i++;
    }

    if (flag == '-') {
        result = -result;
    }

    // 判斷是否越界
    if (result > Integer.MAX_VALUE) {
        return Integer.MAX_VALUE;
    }
    if (result < Integer.MIN_VALUE) {
        return Integer.MIN_VALUE;
    }

    return (int) result;
}