如何避免java中重复的代码？

I'm just beginner in programming. uf is a union-find class with the method union which connects the roots of two node. This piece of code is responsible for opening a site of a grid and union the site with its neighbor if any of neighbor is opened. And if one of its neighbors is full, then fill all nodes that connected with the site. This is the actual code:

我只是编程的初学者。 uf是一个union-find类,其方法union连接两个节点的根。如果打开任何邻居,这段代码负责打开网格的站点并将站点与其邻居联合。如果其中一个邻居已满,则填写与该站点连接的所有节点。这是实际的代码:

        if(i == 1){
            uf.union(len*len, xyTo1D(i,j));
            if(existAndOpen(i+1,j)){
                uf2.union(xyTo1D(i+1,j), xyTo1D(i,j));
                uf.union(xyTo1D(i,j), xyTo1D(i+1,j));
            }
            if(existAndOpen(i-1,j)){
                uf2.union(xyTo1D(i-1,j), xyTo1D(i,j));
                uf.union(xyTo1D(i,j), xyTo1D(i-1,j));
            }
            if(existAndOpen(i,j-1)){
                uf2.union(xyTo1D(i,j-1), xyTo1D(i,j));
                uf.union(xyTo1D(i,j), xyTo1D(i,j-1));
            }
            if(!(j == len && i == len)){
                if(existAndOpen(i,j+1)){
                    uf2.union(xyTo1D(i,j+1), xyTo1D(i,j));
                    uf.union(xyTo1D(i,j), xyTo1D(i,j+1));
                }
            }
        }
        else{
        if(existAndFull(i+1,j)){
            uf2.union(xyTo1D(i+1,j), xyTo1D(i,j));
            uf.union(xyTo1D(i,j), xyTo1D(i+1,j));
        }
        if(existAndFull(i-1,j)){
            uf2.union(xyTo1D(i-1,j), xyTo1D(i,j));
            uf.union(xyTo1D(i,j), xyTo1D(i-1,j));
        }
        if(existAndFull(i,j-1)){
            uf2.union(xyTo1D(i,j-1), xyTo1D(i,j));
            uf.union(xyTo1D(i,j), xyTo1D(i,j-1));
        }
        if(!(j== len && i == len)){
            if(existAndFull(i,j+1)){
                uf2.union(xyTo1D(i,j+1), xyTo1D(i,j));
                uf.union(xyTo1D(i,j), xyTo1D(i,j+1));
            }
        }
        if(existAndOpen(i+1,j)){
            uf.union(xyTo1D(i,j), xyTo1D(i+1,j));
        }
        if(existAndOpen(i-1,j)){
            uf.union(xyTo1D(i,j), xyTo1D(i-1,j));
        }
        if(existAndOpen(i,j-1)){
            uf.union(xyTo1D(i,j), xyTo1D(i,j-1));
        }
        if(!(j== len && i == len)){
            if(existAndOpen(i,j+1)){
                uf.union(xyTo1D(i,j), xyTo1D(i,j+1));
            }
        }
    }
    }

How can I simplify the code?

如何简化代码?

4 个解决方案

#1

Try this

boolean f1(int a, int b) { }
boolean f2(int a, int b) { }
void A(int a, int b) { }

void testAndA(BiPredicate<Integer, Integer> p, int a, int b) {
    if (p.test(a, b))
        A(a, b);
}

and

    if(x == 1){
        testAndA(this::f1, x + 1, y);
        testAndA(this::f1, x, y + 1);
    } else {
        testAndA(this::f2, x + 1, y);
        testAndA(this::f2, x, y + 1);
    }

#2

You can write a loop to loop through all of the different values that could be passed to f1(), such as something like:

您可以编写一个循环来遍历可以传递给f1()的所有不同值,例如:

for (int deltax = -1; deltax <= 1; deltax++) {
    for (int deltay = -1; deltay <= 1; deltay++) {
        if (f1(x + deltax, y + deltay)) {
            A(x + deltax, y + deltay);
        }
    }
}

Of course change the start and end values of deltax and deltay depending on which conditions you need to check.

当然,根据您需要检查的条件,更改deltax和deltay的起始值和结束值。

#3

You say there are "more if statements in each block." Reading between the lines, I assume that means you need to make more calls to f1/f2 and A, but with different x and y offsets.

你说每个街区都有“更多if语句”。在线之间读取,我认为这意味着您需要对f1 / f2和A进行更多调用,但具有不同的x和y偏移。

Here is a program that shows a way to refactor the code to avoid coding repetition. Its main features are:

这是一个程序,它显示了重构代码以避免编码重复的方法。其主要特点是:

It uses object orientation to abstract the inner if blocks.

它使用面向对象来抽象内部if块。

It uses an offsets array to represent the x/y offsets for each inner if block.

它使用偏移数组来表示每个内部if块的x / y偏移量。

The doIt() method uses a loop to invoke the inner if blocks.

doIt()方法使用循环来调用内部if块。

public class Main {

    static interface F {
      void f(int i, int j);
    }

    static class F1Caller implements F {
      public void f(int a, int b) {
        if (f1(a, b)) {
          A(a, b);
        }
      }
    }

    static class F2Caller implements F {
      public void f(int a, int b) {
        if (f2(a, b)) {
          A(a, b);
        }
      }
    }

    static boolean f1(int a, int b) { System.out.print("  f1. "); return true; }
    static boolean f2(int a, int b) { System.out.print("  f2. "); return true; }
    static void A(int a, int b) { System.out.println("a: " + a + ", b: " + b); }

    static F1Caller f1Caller = new F1Caller();
    static F2Caller f2Caller = new F2Caller();

    // x and y offsets for each call to f1/f2.
    // Add more offset rows, as needed.
    static int offsets[][] = {
        {1,  0},
        {0, -1}
    };

    static void doIt(int x, int y) {
      System.out.println("x: " + x + ", y: " + y);

      F f = (x == 1) ? f1Caller : f2Caller;
      for (int k = 0; k < offsets.length; k++) {
          f.f(x + offsets[k][0], y + offsets[k][1]);
      }
    }


    public static void main(String[] args) {
        doIt(0, 0);
        doIt(1, 0);
    }
}

The output of the above program is:

上述程序的输出是:

x: 0, y: 0
  f2. a: 1, b: 0
  f2. a: 0, b: -1
x: 1, y: 0
  f1. a: 2, b: 0
  f1. a: 1, b: -1

#4

You can combine the function together.

您可以将功能组合在一起。

Using a switch.

使用开关。

public static boolean func(int a, int b, int fun)
{
     boolean output = false;
     switch(fun)
     {
     case 1:
         //do stuff
         output = true;
         break;
     case 2:
         //do stuff
         output = true;
         break;
     default:
         //unknown function handling
        output = false;
     }
     return output;
}

Integrate it with a for-loop:

将它与for循环集成:

public static void main(String[] args)
{
     int i;
     //Change this to whatever you want or set it to a argument.
     int repeat = 2;
     for(i = 1; i <= repeat; i++)
     {
          func(a, b, i);
     }
}

#1