确定2D数组中的两个元素是否在同一行或同一列中的最快方法是什么? [英] What is the quickest way to determine if two elements are in the same row or column in a 2D array?

问题描述

作为我的程序之一，我试图检查2D数组中的两个元素是否在同一行或同一列中.我知道可能有很多方法可以解决这个问题，主要是通过循环和遍历数组，但是我只是想知道:有没有一种快速的方法来确定这一点?

As part of one of my programs, I'm trying to check if two elements in a 2D array are part of the same row or column. I know there might be many ways to go about this, mainly with loops and by iterating through the array, but I was just wondering: is there any quick way to determine this?

假设我有一个二维数组，看起来像 {{1，2}，{3，4}，{5，6}} .有没有一种快速的方法来确定 1 和 2 属于同一行?几乎可以在if语句中将其评估为"，如果项目 x 与项目 y 属于同一行，那么就这样"; ?如果没有，那么最快的方法是什么?

Let's say I have a 2D array that looks like {{1, 2}, {3, 4}, {5, 6}}. Would there be any fast way to determine that 1 and 2 belong to the same row? Almost in a way that it could be evaluated in an if statement as "if item x belongs to the same row as item y, then do so and so"? If not, then what would be the quickest way otherwise?

推荐答案

这是一种行处理方法.

int[][] v = {{1, 2}, {3, 4}, {5, 6}};
System.out.println(rowContains(v, 2, 3));
System.out.println(rowContains(v, 5, 6));

打印

false
true

方法

• 流单个d数组，并忽略重复项
• 计数过滤器以查找值.
• 如果最终计数为2，则返回true.

public static boolean rowContains(int[][] v, int v1, int v2) {
    return Arrays.stream(v)
            .map(arrs -> Arrays.stream(arrs)
                    .distinct().filter(a -> a == v1 || a == v2)
                    .count()).anyMatch(a -> a == 2);
}

对于列，最简单的方法是编写一种方法以转置列和行并重新运行该方法.

For columns, the easiest way would be to write a method to transpose the columns and rows and re-run the method.

这篇关于确定2D数组中的两个元素是否在同一行或同一列中的最快方法是什么?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！