使用递归缩进输出答案

【问题标题】：Indenting an output with recursion使用递归缩进输出
【发布时间】：2020-09-22 18:50:06
【问题描述】：

我正在尝试使用递归来打印出两行文本，并且每次出现递归时都会缩进（下面的 javadoc 中的示例）。我知道使用(numCalls + 1) 将允许它每次都增加缩进，但我不确定如何实际将它放在一起，所以它要求的也是。

public class Practice1 {
/**
 * Prints 2 lines of text for each recursive call, indicating call number 
     * (a value >= 1, and <= value of numCalls, as illustrated below.  Each
     * level of recursion should be indicated by indenting the input line by
     * r spaces.  For example, if numCalls is 3, the method should print:
 *  call 1
 *   call 2
 *    call 3
 *    back 3
 *   back 2
 *  back 1
 *  @param r the level of method calls
 *  @param numCalls the number of intended levels
 */

public static void stairSteps(int r, int numCalls) {
    if (r == 0) {
        System.out.println("Done");
    } else {
        System.out.println("call " + r);
        System.out.println("back " + r);
        r--;
        numCalls++;
        stairSteps(r, numCalls);
    }
}

【问题讨论】：

标签： java recursion

【解决方案1】：

public static void stairSteps(int totalLevels, int indent) {
    stairSteps(1, totalLevels, indent);
    System.out.println("Done");
}

private static void stairSteps(int level, int totalLevels, int indent) {
    if (level <= totalLevels) {
        int offs = indent * (level - 1);
        System.out.println(" ".repeat(offs) + "call " + level);
        stairSteps(level + 1, totalLevels, indent);
        System.out.println(" ".repeat(offs) + "back " + level);
    }
}

输出：

stairSteps(3, 2);

call 1
  call 2
    call 3
    back 3
  back 2
back 1
Done

【讨论】：