LeetCode-String to Integer (atoi)

Description：
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. Thefore INT_MIN (−231) is returned.

题意：要求将一个字符串转换为整型，有如下的要求

去除字符串前部的空白符，从第一个非空白符开始转换
注意整型的符号问题，默认为正数，不显示符号‘+’
字符串的末尾可能包含不是数字的内容，需要忽略
如果字符串中不包含合法的关于数字的符号、或者为空，或者第一个字符部位是无关数字的符号，返回值为0
如果转换后的数字超出了int的表示范围，返回值为最大或最小值

解法：首先我们需要去除字符串的前导空白，这样就可以通过首字符判断是否满足条件1和4；要注意的是字符串中可能会包含符号‘+’和‘-’，因此需要记住此时的符号；最后，就是对合法的数字字符进行计算相加，由于字符串的数字位数可能很长，因此用一个double型来存储数字；

class Solution {
    public int myAtoi(String str) {
        if(str == null || str.length() == 0 || str.trim().length() == 0){
            return 0;
        }
        str = str.trim();//去除前导空白和尾部空白
        if(!(str.charAt(0) == '+' || str.charAt(0) == '-' || (str.charAt(0) >= '0' && str.charAt(0) <= '9'))){
            return 0;
        }//首字符为不与数字相关的字符
        int index = 0;//字符串下标
        int symbol = 1;//数字的符号,默认为正数
        double result = 0;//转换后的整数
        if(str.charAt(0) == '-'){
            symbol = -1;
            index++;
        }
        else if(str.charAt(0) == '+'){
            index++;
        }
        while(index < str.length() && str.charAt(index) >= '0' && str.charAt(index) <= '9'){
            result = result * 10 + (str.charAt(index++) - '0');
        }
        result *= symbol;
        if(result > Integer.MAX_VALUE) return Integer.MAX_VALUE;
        if(result < Integer.MIN_VALUE) return Integer.MIN_VALUE;
        return (int)result;
    }
}

LeetCode-String to Integer (atoi)

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