(Leet Code c++)String to Integer(atoi)

8. String to Integer (atoi)

 

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++'s atoi function).

The algorithm for myAtoi(string s) is as follows:

1. Read in and ignore any leading whitespace.
2. Check if the next character (if not already at the end of the string) is '-' or '+'. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
3. Read in next the characters until the next non-digit charcter or the end of the input is reached. The rest of the string is ignored.
4. Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).
5. If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1.
6. Return the integer as the final result.

Note:

• Only the space character ' ' is considered a whitespace character.
• Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.

 

Example 1:

Input: s = "42" Output: 42 Explanation: The underlined characters are what is read in, the caret is the current reader position. Step 1: "42" (no characters read because there is no leading whitespace) ^ Step 2: "42" (no characters read because there is neither a '-' nor '+') ^ Step 3: "42" ("42" is read in) ^ The parsed integer is 42. Since 42 is in the range [-231, 231 - 1], the final result is 42.

Example 2:

Input: s = " -42" Output: -42 Explanation: Step 1: " -42" (leading whitespace is read and ignored) ^ Step 2: " -42" ('-' is read, so the result should be negative) ^ Step 3: " -42" ("42" is read in) ^ The parsed integer is -42. Since -42 is in the range [-231, 231 - 1], the final result is -42.

Example 3:

Input: s = "4193 with words" Output: 4193 Explanation: Step 1: "4193 with words" (no characters read because there is no leading whitespace) ^ Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+') ^ Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit) ^ The parsed integer is 4193. Since 4193 is in the range [-231, 231 - 1], the final result is 4193.

Example 4:

Input: s = "words and 987" Output: 0 Explanation: Step 1: "words and 987" (no characters read because there is no leading whitespace) ^ Step 2: "words and 987" (no characters read because there is neither a '-' nor '+') ^ Step 3: "words and 987" (reading stops immediately because there is a non-digit 'w') ^ The parsed integer is 0 because no digits were read. Since 0 is in the range [-231, 231 - 1], the final result is 0.

Example 5:

Input: s = "-91283472332" Output: -2147483648 Explanation: Step 1: "-91283472332" (no characters read because there is no leading whitespace) ^ Step 2: "-91283472332" ('-' is read, so the result should be negative) ^ Step 3: "-91283472332" ("91283472332" is read in) ^ The parsed integer is -91283472332. Since -91283472332 is less than the lower bound of the range [-231, 231 - 1], the final result is clamped to -231 = -2147483648.

 

Constraints:

• 0 <= s.length <= 200
• s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+', '-', and '.'.

 

문자열을 int형으로 바꾸는 구현문제.

 

주어진 조건을 차례대로 수행하면 된다.

 

처음 시작할 작업은 띄어쓰기를 skip하면 된다.

주어진 s를 한 문자씩 읽으면서 ' '이 아닐 때까지 index를 늘려준다.

 

이후 for문을 index ~ s.size()만큼 반복하여, 문자를 읽도록 하자.

우선 '-'와 '+'가 있는지 확인하자.

처음 시작할 때 sign = 1로 했으니 양수인 상태이다.

그러나 s에서 '-'나 '+'가 있으면 이를 알맞게 바꿔줘야 한다.

 

op 변수는 이전에 '-' 또는 '+'를 읽은 적이 있는지 확인하는 변수다.

num 변수는 answer 문자열에 숫자가 들어있는지 확인하는 변수다.

두 변수 모두 false인 상태여야 부호를 바꿀 수 있다.

만약 둘 중 하나라도 true라면 문자 읽는 것을 멈춰야 한다.

 

다음은 숫자를 읽어야 한다.

'0' ~ '9'에 해당하는 문자가 읽히면 num을 true로 바꿔준다.

이후 answer가 비어있는지 확인한다.

answer가 비어있지 않으면 현재 문자와 answer가 INT_MIN ~ INT_MAX 사이인지 확인해야 한다.

만약 answer + s[i]가 INT_MIN보다 작게 된다면 INT_MIN을 리턴,

INT_MAX보다 크게 된다면 INT_MAX를 리턴하면 된다.

그렇지 않다면 answer에 그대로 현재 숫자를 이어 붙여주도록 한다.

 

이 외의 조건이 나오게 된다면 문자 읽는 것을 멈추도록 한다.

그리고 answer가 비어있는지 확인하여 알맞게 리턴을 해주면 된다.

class Solution {
public:
    int myAtoi(string s) {
        string answer = "";
        int index = 0;
        int length = s.size();
        int sign = 1;
        bool op = false;
        bool num = false;
        
        while(index < length && s[index] == ' '){
            index++;
        }
        
        for(int i = index; i < length; i++){
            if(s[i] == '-' || s[i] == '+'){
                if(!op && !num){
                    op = true;
                    
                    if(s[i] == '-'){
                        sign = -1;
                    }
                }
                else{
                    break;
                }
            }
            else if(s[i] >= '0' && s[i] <= '9'){
                num = true;
                
                if(!answer.empty()){
                    int tmp = sign * stoi(answer);
                    
                    if(tmp > INT_MAX / 10 || tmp == INT_MAX / 10 && (s[i] - '0') > 7){
                        return INT_MAX;
                    }

                    if(tmp < INT_MIN / 10 || tmp == INT_MIN / 10 && (s[i] - '0') > 7){
                        return INT_MIN;
                    }
                }
                answer += s[i];
            }
            else{
                break;
            }
        }
        
        if(!answer.empty()){
            return sign * stoi(answer);
        }
        else{
            return 0;
        }
    }
};