递归一

题目描述

请把纸条竖着放在桌⼦上，然后从纸条的下边向上⽅对折，压出折痕后再展 开。此时有1条折痕，突起的⽅向指向纸条的背⾯，这条折痕叫做“下”折痕 ；突起的⽅向指向纸条正⾯的折痕叫做“上”折痕。如果每次都从下边向上⽅ 对折，对折N次。请从上到下计算出所有折痕的⽅向。

给定折的次数n,请返回从上到下的折痕的数组，若为下折痕则对应元素为"down",若为上折痕则为"up".

测试样例：

1

返回：["down"]

import java.util.Scanner;
 
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
while(sc.hasNextInt()) {
     
 
        int n = sc.nextInt();
        FoldPaper paper=new FoldPaper();
        String s = paper.fold(1, n, true);
        char[] c = s.toCharArray();
        StringBuffer sb = new StringBuffer("[");
        for (int i = 0; i < c.length; i++) {
            if (c[i] == '0') {
                sb.append("\"" + "up" + "\"").append(",");
            } else {
                sb.append("\"" + "down" + "\"").append(",");
            }
        }
        System.out.println(sb.substring(0, sb.length() - 1) + "]");// sb.length()-1，去掉"，"
}
        sc.close();
    }
}
    class FoldPaper {
        public StringBuffer sb = new StringBuffer();
 
        public String fold(int index, int n, boolean isDown) {
 
            if (index > n) {
                return null;
            }
            fold(index + 1, n, true);
            if (isDown) {
                sb.append("1");
            } else {
                sb.append("0");
            }
            fold(index + 1, n, false);
            return sb.toString();
 
        }
 
    }