# (LeetCode) String to Integer (atoi)

## String to Integer (atoi)

• 분류 : String
• 난이도 : Easy

### Problem

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 `[-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.

#### Constraints

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

### Solution

Exapnder

#### Point of Thinking

• 문제에서 주어진 그대로 처리하면 된다.
1. Read in and ignore any leading whitespace.
• `trim`을 이용해서 선행 공백을 제거한다.
1. 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.
• 문자열의 첫 번째 문자를 검증해서 `+`이거나 `-`이면 결과값에 곱해서 반환해주고, 아무것도 없으면 `+` 취급한다. 단 문자열이므로 `+`, `-`가 있으면 인덱스를 올려줘야 한다.
1. 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.
• 2번 스텝 다음 문자열이 `isDigit()`이 아니면 현재까지 계산한 값을 즉시 반환한다.
1. 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).
• String을 Int로 변환하는 `toInt()`에서 해소한다.
1. 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`.
• 문자열 검증시 `Int.MAX_VALUE``Int.MIN_VALUE`에 대한 예외처리만 추가해주면 된다.
1. Return the integer as the final result.
• 여태까지 구한 값을 반환하면 끝