java - 如果有多个结果，则数组的最低值

Question

我需要找到数组中的最小值，但我想知道如何处理多个结果。假设我的数组包含[1,4,7,5,3,1]- 我的结果应该是1,1. 有任何想法吗？

double minimum = array1[0]; //sets the first to be the smallest
for (int i = 0; i < array1.length; i++) //goes through your array
{
 if (array1[i] < array1[0]) //checks and replaces if necessary
 {
    minimum = array[i];   

 }
}

System.out.println( minimum ); //returns the value of the smallest

score 1 · Accepted Answer

1

一件事可以是按升序“排序”您的数组，然后从开始显示值直到它们相等

于 2013-09-17T06:12:31.883 回答

score 1 · Accepted Answer

Arrays.sort(array1);
ArrayList<Integer> smallestValues = new ArrayList<Integer>();
smallestValues.add(array1[0]);
int i=1;
while (i<array1.length && (array1[i] == array1[i-1])) {
     smallestValues.add(array1[i]);
    i++;
 }

score 1 · Accepted Answer

您的代码中有一个小错误，您应该将当前值与最小值进行比较，而不是与第一个值进行比较

double minimum = array1[0]; //sets the first to be the smallest
var minValueCounter = 0;
for (int i = 0; i < array1.length; i++) //goes through your array
{
 if (array1[i] < minimum) //checks and replaces if necessary
 {
    minimum = array[i];  
    minValueCounter  = 1; 

 }
 else if (array1[i] == minimum) //checks and replaces if necessary
 {
    minValueCounter++;

 }
}

score 0 · Accepted Answer

保留另一个变量来计算重复项。如果您找到等于当前最小值的值，请增加此计数。记住每次最小更改时都要重置计数。

score 0 · Accepted Answer

怎么用Collections.sort()

我假设你array1[]的double类型

double array1[] = new double[] { 1, 4, 7, 5, 3, 1 };
ArrayList<Double> al = new ArrayList<Double>();
for (int i = 0; i < array1.length; i++)
    al.add(array1[i]);
Collections.sort(al);
System.out.println(al.toString());

输出：

[1.0, 1.0, 3.0, 4.0, 5.0, 7.0]

要打印所有相等的最小值，请使用

for (int i = 0; i < (al.size() - 1); i++) { // why (al.size() - 1), Its better if try to learn yourself
    if (Double.compare(al.get(i), al.get(i+1))==0)
       System.out.print(""+al.get(i) + "," + al.get(i + 1));
    else
       break;
}

输出：

1.0,1.0

score 0 · Accepted Answer

我会这样做

    int[] a = { 1, 4, 7, 5, 3, 1 };
    Arrays.sort(a);
    int n = 1;
    for (int i = 1; i < a.length && a[i] == a[0]; i++)
        n++;
    int[] res = Arrays.copyOf(a, n);

score 0 · Accepted Answer

只是为了好玩，没有for循环的解决方案：

    Integer[] inputA = new Integer[]{1,4,7,5,3,1};
    List<Integer> inputL = Arrays.asList(inputA);        

    Collections.sort(inputL);

    int last = Collections.frequency(inputL, Collections.min(inputL));

    inputA = Arrays.copyOfRange(inputL.toArray(new Integer[inputL.size()]), 0, last);

    System.out.println(Arrays.deepToString(inputA));

输出：

   [1, 1]

java - 如果有多个结果，则数组的最低值

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