英文:
How should I reduce repetitive code that are all the same but run different functions?
问题
你可以通过创建一个方法，该方法接受一个算法作为参数，然后在不同的循环中调用该方法，并将不同的算法传递给它。这将减少代码的重复性。以下是示例代码：
// 创建一个方法，接受算法作为参数，并返回平均时间
public static long runAlgorithm(YourClass x, int[] randomArray, int algorithmIndex) {
long total = 0;
for (int j = 0; j < 5000; j++) {
if (algorithmIndex == 1) {
x.algorithm1(randomArray[algorithmIndex]);
} else if (algorithmIndex == 2) {
x.algorithm2(randomArray[algorithmIndex]);
} // 可以根据需要添加更多算法
long t2 = System.nanoTime();
long elapsed = t2 - t1;
total += elapsed;
}
return total / 5000;
}
// 在循环中调用方法，并将不同的算法索引传递给它
for (int i = 0; i < 19; i++) {
long t1 = System.nanoTime();
matrix[i][0] = runAlgorithm(x, randomArray, 1); // 调用算法1
}
// 在另一个循环中调用方法，传递不同的算法索引
for (int i = 0; i < 19; i++) {
long t1 = System.nanoTime();
matrix[i][1] = runAlgorithm(x, randomArray, 2); // 调用算法2
}
// 可以继续在其他循环中调用方法，传递不同的算法索引
通过这种方式，你可以避免重复的代码，并轻松更改要调用的算法。你只需要更新 runAlgorithm 方法的条件语句来处理更多的算法选项。
英文:
for (int i = 0; i < 19; i++) {
long t1 = System.nanoTime();
long total = 0;
for (int j = 0; j < 5000; j++) {
x.algorithm1(randomArray[i]);
long t2 = System.nanoTime();
long elapsed = t2 - t1;
total += elapsed;
}
long averageTime = 0;
averageTime= total / 5000;
matrix[i][0] = averageTime;
}
/* Running Algorithm 2 */
for (int i = 0; i < 19; i++) {
long t1 = System.nanoTime();
long total = 0;
for (int j = 0; j < 5000; j++) {
x.algorithm2(randomArray[i]);
long t2 = System.nanoTime();
long elapsed = t2 - t1;
total += elapsed;
}
long averageTime = 0;
averageTime = total / 5000;
matrix[i][1] = averageTime;
}
I have 3 of these for loops that basically do the same thing however the line "x.algorithm2(randomArray[i]);" changes to a different algorithm in each different for loop. How can I reduce this repetitive code to only change the algorithm being called?
答案1
得分: 1
这里是使用函数式接口和泛型的示例代码：
public static <T> void runAlgorithm(long[][] matrix, Consumer<T> tConsumer, T[] tArray) {
for (int i = 0; i < 19; i++) {
long t1 = System.nanoTime();
long total = 0;
for (int j = 0; j < 5000; j++) {
tConsumer.accept(tArray[i]);
long t2 = System.nanoTime();
long elapsed = t2 - t1;
total += elapsed;
}
long averageTime = total / 5000;
matrix[i][1] = averageTime;
}
}
public static void main(String[] args) {
long[][] matrix1 = null; // ???
Integer[] randomInts = new Integer[] {}; // ???
runAlgorithm(matrix1, Algorithms::algorithmOnInts, randomInts);
long[][] matrix2 = null; // ???
Object[] randomObjects = new Object[] {}; // ???
runAlgorithm(matrix2, Algorithms::algorithmOnObjects, randomObjects);
}
public static class Algorithms {
public static void algorithmOnInts(Integer i) {}
public static void algorithmOnObjects(Object o) {}
}
注意：这里的 ??? 部分是示例代码中的占位符，需要您根据实际情况填写相应的内容。
英文:
Here's an example of what you could do using functional interfaces and generics:
public static <T> void runAlgorithm(long[][] matrix, Consumer<T> tConsumer, T[] tArray) {
for (int i = 0; i < 19; i++) {
long t1 = System.nanoTime();
long total = 0;
for (int j = 0; j < 5000; j++) {
tConsumer.accept(tArray[i]);
long t2 = System.nanoTime();
long elapsed = t2 - t1;
total += elapsed;
}
long averageTime = 0;
averageTime = total / 5000;
matrix[i][1] = averageTime;
}
}
public static void main(String[] args) {
long[][] matrix1 = null; // ???
Integer[] randomInts = new Integer[] {}; // ???
runAlgorithm(matrix1, Algorithms::algorithmOnInts, randomInts);
long[][] matrix2 = null; // ???
Object[] randomObjects = new Object[] {}; // ???
runAlgorithm(matrix2, Algorithms::algorithmOnObjects, randomObjects);
}
public static class Algorithms {
public static void algorithmOnInts(Integer i) {}
public static void algorithmOnObjects(Object o) {}
}
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