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Leetcode String to Integer (atoi) problem solution

YASH PAL, 31 July 2024

In this Leetcode String to Integer (atoi) problem solution we need to implement the myAtoi(string s) function that converts a string to a 32 bit signed integer.

Leetcode String to Integer (atoi) problem solution

Problem solution in Python.

class Solution:
    def validCharacter(self, ch):
        return ord('0') <= ord(ch) <= ord('9')
    
    def myAtoi(self, s: str) -> int:
        l = list(s)
        
        # remove whitespace
        while l and l[0] == " ":
            l.pop(0)
            
        # specify sign
        sign = 1
        if l and l[0] == '+':
            l.pop(0)
            sign = 1
        elif l and l[0] == '-':
            l.pop(0)
            sign = -1
        
        buf = ""
        while l and self.validCharacter(l[0]):
            buf += l.pop(0)
        
        result = 0
        for i in buf:
            num = ord(i) - ord('0')
            result = result* 10 + num
        result *= sign  # add sign
        if result > 2 ** 31 - 1:
            return 2 ** 31 - 1
        elif result < -(2 ** 31):
            return -(2 ** 31)
        else:
            return result

Problem solution in Java.

int start = 0, sign = 1;
        long res = 0;
        while(start < s.length() && s.charAt(start) == ' ') {
            start++;
        }
        
        if(start < s.length() && s.charAt(start) == '+') {
            start++;
        } else if(start < s.length() && s.charAt(start) == '-') {
            sign = -1;
            start++;
        }
        
        while(start < s.length()) {
            if(!Character.isDigit(s.charAt(start))) break;
            res *= 10;
            res += s.charAt(start) - '0';
            if(res > Integer.MAX_VALUE) {
                if(sign == 1) return Integer.MAX_VALUE;
                return Integer.MIN_VALUE;
            }
            start++;
        }
        
        return (int) res * sign;

Problem solution in C++.

class Solution {
public:
    int myAtoi(string str) {
        long long ans =0;
        bool isNeg = false;
        int i=0;
        while(str[i]==' ')
            i++;
        
        if(str[i]=='-'){
            isNeg = true;
            i++;
        }
            
        else if(str[i]=='+'){
            isNeg = false;
            i++;
        }
        while(str[i]=='0')
            i++;
        
        int end=0;
        for( end=i;end<str.size();end++){
             if(str[end]>'9'||str[end]<'0')
            break;
        }
        if(end-i>12){
            if(isNeg)
                return INT_MIN;
            else return INT_MAX;
        }
        for(int j=i;j<str.size();j++){
          if(str[j]>'9'||str[j]<'0')
            break;
            
            ans = 10*ans + str[j]-'0';
           
        }
         if(isNeg)
                return (INT_MIN>-ans?INT_MIN:-ans);
            else
                return (INT_MAX>ans?ans:INT_MAX);
       
    }
};

Problem solution in C.

int myAtoi(char *str) {
    int flag = 1;
    long res = 0;
    
    if (str == NULL) return 0;
    while (*str == ' ' || *str == 't') ++str;
    if (str == NULL) return 0;
    if (*str != '-' && *str != '+' && (*str < '0' || *str >'9')) return 0; 
    if (*str == '-') {flag = -1;++str;}
    else if (*str == '+') {flag = 1; ++str;}
    if (str == NULL) return 0;
    
    int cnt = 0;  
    while(*str) {
        if (*str < '0' || *str >'9') break;
        res = res * 10 + *str - 0x30;
        ++str;
        ++cnt;
    }
    
    if (cnt > 10) return flag == 1 ? INT_MAX : INT_MIN;
    
    res *= flag;
    if (res > INT_MAX) return INT_MAX;
    if (res < INT_MIN) return INT_MIN;
    

    return (int)res;
}

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