22. 문자열 압축

https://programmers.co.kr/learn/courses/30/lessons/60057

문자열 압축
2020 KAKAO BLIND RECRUITMENT
4461명 완료

코딩테스트 연습 - 문자열 압축

파이썬

def solution(s):
    min_length = []
    for length in range(1, len(s) + 1):
        temp, compressed, count = '', '', 0
        for i in range(0, len(s), length):
            word = s[i: i + length]
            if word == temp:
                count += 1
            else:
                compressed += (str(count) if (count > 1) else '') + temp
                count = 1
                temp = word
        compressed += (str(count) if (count > 1) else '') + temp
        min_length.append(len(compressed))
    return min(min_length)

JS

function solution(s) {
    const min_length = []
    for (let i = 1; i <= s.length; i++) {
        let temp = ''
        let compressed = ''
        let count = 0
        for (let j = 0; j < s.length; j += i) {
            let word = s.slice(j, j + i)
            if (temp === word)
                count++
            else {
                compressed += ((count > 1 ? count : '') + temp)
                count = 1
                temp = word
            }
        }
        compressed += ((count > 1 ? count : '') + temp)
        min_length.push(compressed.length)
    }
    return min_length.reduce((a, b) => (a > b) ? b : a)
}

Java

class Solution {
    public int solution(String s) {
        int answer = 1000;
        for (var length = 1; length <= s.length(); length++) {
            var count = 0;
            var lastStr = "";
            var compressedStr = new StringBuilder();
            for (var index = 0; index < s.length(); index += length) {
                var nowStr = s.substring(index, Math.min(index + length, s.length()));
                if (lastStr.equals(nowStr)) {
                    count++;
                } else { // 문자열이 바뀜
                    if (count > 1) compressedStr.append(count);
                    compressedStr.append(lastStr);
                    count = 1;
                    lastStr = nowStr;
                }
            }
            if (count > 1) compressedStr.append(count);
            compressedStr.append(lastStr);
            answer = Math.min(compressedStr.length(), answer);
        }
        return answer;
    }
}