String to Integer (atoi)

最新推荐文章于 2019-08-31 13:11:50 发布

转载最新推荐文章于 2019-08-31 13:11:50 发布 · 64 阅读

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原文链接：http://www.cnblogs.com/pilihaotian/p/5030741.html

本文介绍了一个将字符串转换为整数（atoi）的方法，并详细解释了如何处理各种输入情况，包括空字符串、空白字符、正负号以及数值计算等。

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.

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.

Analysis

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

public class Solution {
    public int myAtoi(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;
    }
}

转载于:https://www.cnblogs.com/pilihaotian/p/5030741.html