#백준_1406 에디터

유형 : 링크드 리스트

 

# 처음에는 입력받아서 arrayList를 사용하고 index를 조절하며 switch case 문으로 해결하려 했으나 시간초과로 문제를 풀 수 없었다. 다른 방법을 찾아보던 중 스택으로 해결할 수 있다는 글을 보고 스택으로 수정하여 풀었더니 통과하였다. 시간 제한을 신경써야 함.

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import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.StringTokenizer;   public class Main {     public static void main(String[] args) throws IOException {         BufferedReader br = new BufferedReader(new InputStreamReader(System.in));         String str =br.readLine();                  ArrayList<Character> arrList = new ArrayList<>();         for(int i=0; i<str.length(); i++) {             arrList.add(str.charAt(i));         }                  int N = Integer.parseInt(br.readLine());                  int index = arrList.size();         for(int i=0; i<N; i++) {             StringTokenizer st = new StringTokenizer(br.readLine());             char mod = st.nextToken().charAt(0);             switch (mod) {             case 'L':                 if(index==0)                     break;                 else {                     index-=1;                     break;                     }             case 'D':                 if(arrList.size()==0 || arrList.size()==index)                     break;                 else {                     index+=1;                     break;                 }             case 'B':                 if(index==0 || arrList.size()==0)                     break;                 else {                     arrList.remove(index-1);                     index--;                     break;                 }                                               case 'P':                 char c = st.nextToken().charAt(0);                 arrList.add(index,c);                 index++;                 break;             } //            for(int q=0; q<arrList.size(); q++) { //                System.out.print(arrList.get(q)); ////            }System.out.println();         }             for(int q=0; q<arrList.size(); q++) {                 System.out.print(arrList.get(q));             }System.out.println();              } }   Colored by Color Scripter

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시간 초과남, 따라서 선형 리스트인 스택을 사용하여 문제를 접근하였고 해결하였음

 

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package bj;   import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.Stack; import java.util.StringTokenizer; public class p1406 {      public static void main(String[] args) throws IOException {                    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));          String str = br.readLine();                    int n = Integer.parseInt(br.readLine());          Stack<Character> st1 = new Stack<>();          Stack<Character> st2 = new Stack<>();                    for(int i=0; i<str.length(); i++) {              st1.add(str.charAt(i));          }                    for(int i=0; i<n; i++) {              StringTokenizer st = new StringTokenizer(br.readLine());              char ch = st.nextToken().charAt(0);                            if(ch == 'L') {                 if(!st1.isEmpty()) {                     char c = st1.pop();                     st2.push(c);                 }              }else if(ch == 'D') {                  if(!st2.isEmpty()) {                      char c = st2.pop();                      st1.push(c);                  }              }else if(ch =='B') {                  if(!st1.isEmpty())                      st1.pop();                                }else if(ch == 'P') {                  char tmp = st.nextToken().charAt(0);                  st1.push(tmp);              }          }                    while(!st1.isEmpty())              st2.push(st1.pop());                    while(!st2.isEmpty()) {              System.out.print(st2.pop());          }                          } }   Colored by Color Scripter

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