通过递归找到数组中最大的正整数 [英] Finding the largest positive int in an array by recursion

问题描述

我决定递归地实现一个非常简单的程序，看看 Java 处理递归的效果如何*，结果有点简短.这就是我最终写的:

I decided to implement a very simple program recursively, to see how well Java handles recursion*, and came up a bit short. This is what I ended up writing:

public class largestInIntArray {
  public static void main(String[] args)
  {
    // These three lines just set up an array of ints:
    int[] ints = new int[100];
    java.util.Random r = new java.util.Random();
    for(int i = 0; i < 100; i++) ints[i] = r.nextInt();

    System.out.print("Normal:"+normal(ints,-1)+" Recursive:"+recursive(ints,-1));
  }

  private static int normal(int[] input, int largest) {
    for(int i : input)
      if(i > largest) largest = i;

    return largest;
  }

  private static int recursive(int[] ints, int largest) {
    if(ints.length == 1)
      return ints[0] > largest ? ints[0] : largest;

    int[] newints = new int[ints.length - 1];
    System.arraycopy(ints, 1, newints, 0, ints.length - 1); 

    return recursive(newints, ints[0] > largest ? ints[0] : largest);
  }
}

这很好用，但由于它有点难看，我想知道是否有更好的方法.如果有人有任何想法/替代方案/语法糖可以分享，那将不胜感激！

And that works fine, but as it's a bit ugly I wondered if there was a better way. If anyone has any thoughts/alternatives/syntactic sugar to share, that'd be much appreciated!

附言如果你说使用 Lisp"，你什么也得不到(但尊重).我想知道这是否可以在 Java 中看起来不错.

P.s. If you say "use Lisp" you win nothing (but respect). I want to know if this can be made to look nice in Java.

*以及我如何处理递归

推荐答案

2 项改进:

• 没有数组的副本(仅使用偏移量)

• 无需给出当前最大值

• no copy of the array (just using the offset)

• no need to give the current max

private static int recursive(int[] ints, int offset) {
    if (ints.length - 1 == offset) {
        return ints[offset];
    } else {
        return Math.max(ints[offset], recursive(ints, offset + 1));
    }
}

以 recursive(ints, 0) 开始递归.

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