java如何使用单个for循环执行多个数组操作？

Question

所以我被指示做下面的事情，我可以让它运行得很好，但是它只需要一个for循环，我就是不知道怎么做。任何帮助都将不胜感激。这就是我想到的。我需要能够得到总数，平均数，最高数和最低数与一个循环

public static void arrayTotalAndAverage(int[] array) {

        int[] numbers = { 10, 4, 13, 29, 57, 92, 114, 212, 3, 88, 36, 101, 77, 42, 209 };

        int total = 0;
        double average = 0;
        int highest = numbers[0];
        int lowest = numbers[0];

        for (int i = 0; i < numbers.length; i++) {
            total += numbers[i];
            average = total / numbers.length;
        }

        for (int i = 1; i < numbers.length; i++) {
            if (numbers[i] > highest)
                highest = numbers[i];
            if (numbers[i] < lowest)
                lowest = numbers[i];
        }

        System.out.println("Total");
        System.out.println(total);

        System.out.println("Average");
        System.out.println(average);

        System.out.println("Highest number");
        System.out.println(highest);

        System.out.println("Lowest number");
        System.out.println(lowest);

    }

Answer 1

你可以像这样在一个循环中完成所有的操作。平均值可以从循环中移出

int[] numbers = { 10, 4, 13, 29, 57, 92, 114, 212, 3, 88, 36, 101, 77, 42, 209 };

int total = 0;
double average = 0;
int highest = Integer.MIN_VALUE;
int lowest = Integer.MAX_VALUE;

for (int i = 0; i < numbers.length; i++) {
    total += numbers[i];
    if (numbers[i] > highest)
        highest = numbers[i];
    if (numbers[i] < lowest)
        lowest = numbers[i];
}
average = total / numbers.length;

System.out.println("Total");
System.out.println(total);

System.out.println("Average");
System.out.println(average);

System.out.println("Highest number");
System.out.println(highest);

System.out.println("Lowest number");
System.out.println(lowest);

}

Answer 2

只需将第二个循环与第一个循环合并即可。请注意，应该只在循环结束时计算平均值，并且必须首先将其中一个操作数转换为double，以避免整数除法的截断

int total = 0;
double average = 0;
int highest = numbers[0];
int lowest = numbers[0];

for (int i = 0; i < numbers.length; i++) {
 total += numbers[i];
 if (numbers[i] > highest)
  highest = numbers[i];
 if (numbers[i] < lowest)
  lowest = numbers[i];
}
average = (double) total / numbers.length;

Demo

输出：

Total
1087
Average
72.46666666666667
Highest number
212
Lowest number
3

Answer 3

您可以通过StreamAPI完成所有这些操作

 int[] numbers = {10, 4, 13, 29, 57, 92, 114, 212, 3, 88, 36, 101, 77, 42, 209};
    IntSummaryStatistics statistics = IntStream.of(numbers).summaryStatistics();
    
    System.out.println("total: " + statistics.getSum());
    System.out.println("average: " + statistics.getAverage());
    System.out.println("minimum: " + statistics.getMin());
    System.out.println("maximum: " + statistics.getMax());

Answer 4

实际上，在这里进行操作的最有效方法可能是只迭代数组一次，然后跟踪最小值、最大值、平均值和总值的状态：

int[] numbers = { 10, 4, 13, 29, 57, 92, 114, 212, 3, 88, 36, 101, 77, 42, 209 };
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
int total = 0;
double average = 0;

for (int i=0; i < numbers.length; ++i) {
    total += numbers[i];
    if (numbers[i] > max) {
        max = numbers[i];
    }
    if (numbers[i] < min) {
        min = numbers[i];
    }
}

average = 1.0d * total / numbers.length;

System.out.println("total: " + total);
System.out.println("average: " + average);
System.out.println("minimum: " + min);
System.out.println("maximum: " + max);

这张照片是：

total: 1087
average: 72.46666666666667
minimum: 3
maximum: 212