8 String to Integer (atoi)

Posted 2022-04-17 xiaoyisun06

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.

这道题真是太坑了，强烈不推荐。不过考察的知识点还是不错的，主要是char 类型变量和ASCII 码的一些操作，还有数值溢出的考察，不过它要考虑的各种特殊情况真是太多了，总之很烦。

class Solution 
public:
    int myAtoi(string str) 
        
        vector<int> num;
        bool sign = true;
        
        long r = 0;
        int flag = 0;
        
        for(int i=0;i<str.length();i++)
        
            char curr = str[i];
            
            if(curr == ' '&& flag==0) continue;
            else if(curr == '-' && flag == 0) 
            
                sign = false;
                flag++;
                continue;
            
            else if(curr == '+'&& flag == 0) 
            
                sign = true;
                flag++;
                continue;
            
            else if((curr >= '0')&& (curr<= '9')) 
            
                if(r== 0) flag++;
                
                r = r*10 + int(curr-'0');
                
                if(sign&&(r > INT_MAX))
                
                    r= INT_MAX;
                    break;
                    
                if( (!sign) && ((-r)<INT_MIN))
                
                    r=INT_MIN;
                    break;
                
            
            
            else break;
                
        
        
        if((sign == false) && (r!=INT_MIN)) r = 0-r;
        
        return r;
        
    
;