Lv.0 문자열 여러 번 뒤집기

문제설명

문자열 my_string과 이차원 정수 배열 queries가 매개변수로 주어집니다. queries의 원소는 [s, e] 형태로, my_string의 인덱스 s부터 인덱스 e까지를 뒤집으라는 의미입니다. my_string에 queries의 명령을 순서대로 처리한 후의 문자열을 return 하는 solution 함수를 작성해 주세요.

 

제한사항

my_string은 영소문자로만 이루어져 있습니다.
1 ≤ my_string의 길이 ≤ 1,000
queries의 원소는 [s, e]의 형태로 0 ≤ s ≤ e < my_string의 길이를 만족합니다.
1 ≤ queries의 길이 ≤ 1,000

 

입출력 예

my_string	queries	result
"rermgorpsam"	[[2, 3], [0, 7], [5, 9], [6, 10]]	"programmers"

 

입출력 예 설명

예제 1번의 my_string은 "rermgorpsam"이고 주어진 queries를 순서대로 처리하면 다음과 같습니다.

queries	my_string
	"rermgorpsam"
[2, 3]	"remrgorpsam"
[0, 7]	"progrmersam"
[5, 9]	"prograsremm"
[6, 10]	"programmers"

따라서 "programmers"를 return 합니다.

 

풀이

 

JAVA

class Solution {
    public String solution(String my_string, int[][] queries) {
        char[] arr = my_string.toCharArray(); // 문자열을 문자 배열로 변환
        
        for (int[] query : queries) {
            int start = query[0];
            int end = query[1];
            
            while (start < end) {
                char temp = arr[start];
                arr[start] = arr[end];
                arr[end] = temp;
                
                start++;
                end--;
            }
        }
        
        return new String(arr); 
    }
}

*다른 코드

class Solution {

    char[] arr;

    public String solution(String my_string, int[][] queries) {

        arr = my_string.toCharArray();

        for (int[] query : queries) {
            reverse(query[0], query[1]);
        }

        return new String(arr);
    }

    private void reverse(int s, int e) {
        while (s < e) {
            char temp = arr[s];
            arr[s++] = arr[e];
            arr[e--] = temp;
        }
    }
}

 

Python

def solution(my_string, queries):
    for (s, e) in queries:
        my_string = my_string[:s] + my_string[s:e+1][::-1] + my_string[e+1:]
    return my_string

*다른 코드

def solution(my_string, queries):
    answer=list(my_string)
    for s,e in queries:
        answer[s:e+1]=answer[s:e+1][::-1]
    return ''.join(answer)

 

JS

function solution(my_string, queries) {
    return queries.reduce((acc, [s, e]) => {
        const start = acc.slice(0, s);
        const mid = acc.slice(s, e + 1).split("").reverse().join("");
        const end = acc.slice(e + 1);
        return start + mid + end;
    }, my_string)
}

*다른 코드

function solution(my_string, queries) {
    let str = my_string.split('');
  queries.forEach(([start, end]) => {
    const changeStr = str.slice(start, end + 1);
    str.splice(start, changeStr.length, ...changeStr.reverse());
  });
  return str.join('');
}