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Original file line number Diff line number Diff line change
@@ -0,0 +1,44 @@
package week05.boj13417;

import java.util.*;
import java.io.*;

public class 카드문자열_박나현 {
public static void main(String[] args) throws IOException {
StringBuilder sb = new StringBuilder();

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());


int count = Integer.parseInt(st.nextToken());

while (count-- > 0) {

Deque<String> deq = new ArrayDeque<>();

st = new StringTokenizer(br.readLine());
int length = Integer.parseInt(st.nextToken());

st = new StringTokenizer(br.readLine());

while (length-- > 0) {
String c = st.nextToken();
if (deq.isEmpty()) {
deq.offer(c);
Comment on lines +27 to +28
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여기까지 작성한 뒤에 continue (아마 break 아니고 continue일 것 같은데 잘못됐다면 죄송합니다 ㅜㅜ) 로 끊어주면
조건문 중첩이 줄어서 가독성이 더 좋아질 것 같습니다..!

} else {
if (c.compareTo(deq.peekFirst()) > 0) {
deq.offerLast(c);
} else deq.offerFirst(c);
}
}

while (!deq.isEmpty()) {
sb.append(deq.poll());
}

sb.append("\n");
}
System.out.println(sb);
}
}
Original file line number Diff line number Diff line change
@@ -0,0 +1,53 @@
package week05.boj20301;

import java.io.*;
import java.util.*;

public class 반전요세푸스_박나현 {
static Deque<Integer> deq = new ArrayDeque<>();
static StringBuilder sb = new StringBuilder();
static int count = 0;
static boolean turn = false;

public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());

int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
int m = Integer.parseInt(st.nextToken());

for (int i = 1; i <= n; i++) {
deq.offerLast(i);
}

while (!deq.isEmpty()) {
if (!turn) {
turnRight(k);
} else {
turnLeft(k);
}

count++;
if (count == m) {
count = 0;
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저는 배수로 판별하는 방법 밖에 생각 못 했는데 내부에서 다시 초기화 해주는 방법도 있군요!! 배워갑니다 :)

turn = !turn;
}
}
System.out.println(sb);
}

private static void turnRight(int k) {
for (int i = 0; i < k-1; i++) {
deq.offerLast(deq.pollFirst());
}
sb.append(deq.pollFirst()).append("\n");
}

private static void turnLeft(int k) {
for (int i = 0; i < k-1; i++) {
deq.offerFirst(deq.pollLast());
}
sb.append(deq.pollLast()).append("\n");
}
}
Original file line number Diff line number Diff line change
@@ -0,0 +1,47 @@
package week05.boj2346;

import java.util.*;
import java.io.*;

// 메모리: 15564KB | 시간: 164ms
public class 풍선터뜨리기_박나현 {
public static void main(String[] args) throws IOException {

StringBuilder sb = new StringBuilder();
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

// index 구하는 메서드가 없어 배열로 저장
Deque<int[]> deq = new ArrayDeque<>();
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덱을 배열로 넣어서 푸신 아이디어가 신박하네요 ..!
풍선 객체를 만들어서 푸는 방법과 유사한 것 같습니다 :)


// size 만큼 deq에 값 저장
int size = Integer.parseInt(br.readLine());

StringTokenizer st = new StringTokenizer(br.readLine());
for (int i = 1; i <= size; i++) {
int n = Integer.parseInt(st.nextToken());
deq.offer(new int[]{i, n});
}

int[] first = deq.pollFirst();
sb.append(first[0]).append(" ");

int next = first[1];

while (!deq.isEmpty()) {
if (next > 0) {
for (int j = 0; j < next-1; j++) {
deq.offerLast(deq.poll());
}
} else {
for (int j = 0; j < Math.abs(next); j++) {
deq.offerFirst(deq.pollLast());
}
}

first = deq.poll();
sb.append(first[0]).append(" ");
next = first[1];
}
System.out.println(sb);
}
}