[LeetCode] 8. String to Integer (atoi) (Java)
2023. 1. 3. 07:47ㆍ알고리즘/LeetCode
Description
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:
- Read in and ignore any leading whitespace.
- 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.
- 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.
- 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).
- 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.
- Return the integer as the final result.
문자를 32-bit 정수로 바꾸는 알고리즘을 만들기.
1. 모든 공백을 무시하기.
2. 다음 문자가 '-'인지 '+'인지에 따라 양수인지 음수인지 결정하기.
3. 숫자가 아닌 문자일 때 까지 숫자를 읽고 남은 문자는 무시한다.
4. 읽은 숫자를 정수로 변환한다. 숫자가 읽히지 않았다면 정수 0을 반환한다.
5. 32-bit 정수의 범위를 벗어나면 최대 범위 값을 준다.
6. 결과를 반환한다.
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
class Solution {
public int myAtoi(String s) {
int start = -1, end = -1;
for(int i = 0; i < s.length(); i++){
if(Character.isLetter(s.charAt(i)) || s.charAt(i) == '.'){
break;
}
if(s.charAt(i) == ' '){
continue;
}
if(s.charAt(i) == '-' || s.charAt(i) == '+' || Character.isDigit(s.charAt(i))){
start = i;
end = i;
i+=1;
while(i < s.length() && Character.isDigit(s.charAt(i))){
end=i;
i+=1;
}
break;
}
}
if(start == end && (end == -1 || s.charAt(end) == '-' || s.charAt(end) == '+')){
return 0;
}
int result = 0;
try{
result = Integer.parseInt(s.substring(start, end+1));
} catch (NumberFormatException e){
if(s.charAt(start) == '-'){
return Integer.MIN_VALUE;
}
else{
return Integer.MAX_VALUE;
}
}
return result;
}
}
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