(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.

Example 1

1
2
3
4
5
6
7
8
9
10
11
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 [-2^31, 2^31 - 1], the final result is 42.

Example 2

1
2
3
4
5
6
7
8
9
10
11
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 [-2^31, 2^31 - 1], the final result is -42.

Example 3

1
2
3
4
5
6
7
8
9
10
11
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 [-2^31, 2^31 - 1], the final result is 4193.

Example 4

1
2
3
4
5
6
7
8
9
10
11
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 [-2^31, 2^31 - 1], the final result is 0.

Example 5

1
2
3
4
5
6
7
8
9
10
11
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 '.'.

Solution

Exapnder
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
class Solution {
fun myAtoi(s: String): Int {
if (s.isNullOrBlank()) {
return 0
}
var str = s.trim()
var sign = 1
var index = 0
when (str[0]) {
'-' -> {
sign = -1
index++
}
'+' -> {
sign = 1
index++
}
else -> {
index = 0
}
}

var result = 0L
for (i in index until str.length) {
if (!str[i].isDigit()) {
break
}
result = result * 10 + (str[i] - '0')
if (sign == 1 && result > Int.MAX_VALUE) {
return Int.MAX_VALUE
}
if (sign == -1 && -result < Int.MIN_VALUE) {
return Int.MIN_VALUE
}
}
return result.toInt() * sign
}
}

Point of Thinking

  • 문제에서 주어진 그대로 처리하면 된다.
  1. Read in and ignore any leading whitespace.
  • trim을 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_VALUEInt.MIN_VALUE에 대한 예외처리만 추가해주면 된다.
  1. Return the integer as the final result.
  • 여태까지 구한 값을 반환하면 끝