/* 문자열안에 문자열
* 문자열 str1, str2가 매개변수로 주어집니다.
* str1 안에 str2가 있다면 1을 없다면 2를 return하도록 solution 함수를 완성해주세요.
*
* str1 str2 result
* "ab6CDE443fgh22iJKlmn1o" "6CD" 1
* "ppprrrogrammers" "pppp" 2
*
* "ab6CDE443fgh22iJKlmn1o" str1에 str2가 존재하므로 1을 return합니다.
* "ppprrrogrammers" str1에 str2가 없으므로 2를 return합니다.
*/
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 | public class programmer_0_11 { static String a1 = "ab6CDE443fgh22iJKlmn1o"; static String a2 = "6CD"; static String b1 = "ppprrrogrammers"; static String b2 = "pppp"; public int solution(String str1, String str2){ int answer = 2; //다른사람 풀이 : 삼항연산자 contains 사용 //return (str1.contains(str2)? 1: 2); for(int i=0; i<str1.length(); i++){ int end = 0; if((i+str2.length()) < str1.length()) end = i + str2.length(); else end = str1.length(); if(str1.substring(i,end).equals(str2)){ answer = 1; break; } } return answer; } public static void main(String args[]){ programmer_0_11 t = new programmer_0_11(); System.out.println("---------------------------------------"); System.out.println("result = " + t.solution(a1, a2)); System.out.println("---------------------------------------"); System.out.println("result2 = " + t.solution(b1, b2)); System.out.println("---------------------------------------"); } } | cs |