Java非递归方式获取目录及其子目录中的所有文件

发布于 2024-11-25 17:13:24 字数 389 浏览 1 评论 0原文

我正在尝试获取目录及其子目录中所有文件的列表。我当前的递归方法如下：

private void printFiles(File dir) {
  for (File child : dir.listFiles()) {
    if (child.isDirectory()) {
      printFiles(child);
    } else if (child.isFile()) {
      System.out.println(child.getPath());
    }
  }
}

printFiles(new File("somedir/somedir2"));

但是，我希望有一种非递归方法（可能是现有的 API 调用）来执行此操作。如果不是，这是最干净的方法吗？

原文

I am trying to get a list of all files in a directory and its subdirectories. My current recursive approach is as follows:

private void printFiles(File dir) {
  for (File child : dir.listFiles()) {
    if (child.isDirectory()) {
      printFiles(child);
    } else if (child.isFile()) {
      System.out.println(child.getPath());
    }
  }
}

printFiles(new File("somedir/somedir2"));

However, I was hoping there was a non-recursive way (an existing API call, maybe) of doing this. If not, is this the cleanest way of doing this?

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ぃ双果 2024-12-02 17:13:24

您始终可以使用堆栈（对于 DFS）或队列（对于 BFS）将递归解决方案替换为迭代解决方案：

private void printFiles(File dir) {
  Stack<File> stack = new Stack<File>();
  stack.push(dir);
  while(!stack.isEmpty()) {
    File child = stack.pop();
    if (child.isDirectory()) {
      for(File f : child.listFiles()) stack.push(f);
    } else if (child.isFile()) {
      System.out.println(child.getPath());
    }
  }
}

printFiles(new File("abc/def.ghi"));

You can always replace a recursive solution with an iterative one by using a stack (for DFS) or a Queue (For BFS):

private void printFiles(File dir) {
  Stack<File> stack = new Stack<File>();
  stack.push(dir);
  while(!stack.isEmpty()) {
    File child = stack.pop();
    if (child.isDirectory()) {
      for(File f : child.listFiles()) stack.push(f);
    } else if (child.isFile()) {
      System.out.println(child.getPath());
    }
  }
}

printFiles(new File("abc/def.ghi"));

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