[Leetcode] 8. String to Integer (atoi)

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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 character 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 [-2^31, 2^31 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -2^31 should be clamped to -2^31, and integers greater than 2^31 - 1 should be clamped to 2^31 - 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.

Constraints:

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

Solution

문자열을 읽어서 숫자로 반환하는 함수를 만드는 것이 이번 문제..!

step1 ) whitespace를 넘기고

step2 ) + / - 일때는 index++, 부호 저장

step3) 문자가 나오거나 string s끝까지 읽을 때까지 문자를 저장하는데 이 때 문제 7번풀때처럼 INT_MAX/10을 나누는 방법을 활용했더니 처음으로 제대로 Accept..!

class Solution {
public:
    int myAtoi(string s) {
        int l = s.length();
        int ret = 0;
        int i=0;
        bool negative = false;

        // step 1
        while(isspace(s[i]))
        {
            i++;
        }
        //step 2
        if(s[i]=='-'){
            negative = true;
            i++;
        }
        else if(s[i]=='+'){
            i++;
        }

        //step 3
        while(isdigit(s[i]) && i<l)
        {
            int num = s[i]-'0';
            if(ret>(INT_MAX-num)/10){
                return negative? INT_MIN:INT_MAX;
            }
            // cout << ret << "+" << num << endl;
            ret = ret*10+num;
            i++;
        }
        return negative? ret*(-1): ret;
    }
};

다른 사람들 푼걸 보니

isspace 대신 s[index] == ' '

isdigit 대신 return ch >= '0' && ch <= '9';

integer overflow를 피하는 방법으로는 if(result > (INT_MAX / 10) || (result == (INT_MAX / 10) && digit > 7)){

이렇게 쓴 것을 확인!