8. String to Integer (atoi)

Posted 2022-12-11 Carl_Hugo

https://leetcode.com/problems/string-to-integer-atoi/description/

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.

解题思路：
1.通过pOrn判断符号
2.删除字符串前的空格
3.是数字就累乘相加

public class Solution 

    public int myAtoi(String str) 
        //将字符串转换为字符数组
        char[] array = str.toCharArray();
        double result=0;
        //正数或负数
        int pOrn=1;
        int count=0;
        for(char c:array)
            count++;
            if(c>='0'&&c<='9')
                result=result*10+(c-'0');
            else if(c=='+'&&count==1)
                pOrn=1;
            else if(c=='-'&&count==1)
                pOrn=-1;
            else if(c==' '&&count==1)
                count--;
            else
                break;
            
        

        if(result>Integer.MAX_VALUE)
            if(pOrn==1)
                return Integer.MAX_VALUE;
            else
                return Integer.MIN_VALUE;
            
        else
            return (int)result*pOrn;