如何在没有一个其他数组的情况下制作一个数组？

Question

I am trying to make a code with two arrays. The second array has the same values of the first except for the smallest number. I have already made a code where z is the smallest number. Now I just want to make a new array without z, any feedback would be appreciated.

public static int Second_Tiny() {
    int[] ar = {19, 1, 17, 17, -2};
    int i;
    int z = ar[0];

    for (i = 1; i < ar.length; i++) {
        if (z >ar[i]) {  
            z=ar[i];  
        }                   
    }
}

Answer 1

Java 8 streams have built in functionality that can achieve what you're wanting.

public static void main(String[] args) throws Exception {
    int[] ar = {19, 1, 17, 17, -2, -2, -2, -2, 5};

    // Find the smallest number
    int min = Arrays.stream(ar)
            .min()
            .getAsInt();

    // Make a new array without the smallest number
    int[] newAr = Arrays
            .stream(ar)
            .filter(a -> a > min)
            .toArray();

    // Display the new array
    System.out.println(Arrays.toString(newAr));
}

Results:

[19, 1, 17, 17, 5]

Otherwise, you'd be looking at something like:

public static void main(String[] args) throws Exception {
    int[] ar = {19, 1, 17, 17, -2, -2, -2, -2, 5};

    // Find the smallest number
    // Count how many times the min number appears
    int min = ar[0];
    int minCount = 0;
    for (int a : ar) {
        if (minCount == 0 || a < min) {
            min = a;
            minCount = 1;
        } else if (a == min) {
            minCount++;
        }
    }

    // Make a new array without the smallest number
    int[] newAr = new int[ar.length - minCount];
    int newIndex = 0;
    for (int a : ar) {
        if (a != min) {
            newAr[newIndex] = a;
            newIndex++;
        }
    }

    // Display the new array
    System.out.println(Arrays.toString(newAr));
}

Results:

[19, 1, 17, 17, 5]

Answer 2

I think the OP is on wrong track seeing his this comment:

"I am trying to find out the second smallest integer in array ar[]. I should get an output of 1 once I am done. The way I want to achieve that is by making a new array called newar[] and make it include all the indexes of ar[], except without -2."

This is a very inefficient way to approach this problem. You'll have to do 3 passes, Once to find to smallest indexed element, another pass to remove the element (this is an array so removing an element will require a full pass), and another one to find smallest one again.

You should just do a single pass algorithm and keep track of the smallest two integers, or even better use a tree for efficiency. Here are the best answers of this problem:

Find the 2nd largest element in an array with minimum number of comparisons

Algorithm: Find index of 2nd smallest element from an unknown array

UPDATE: Here is the algorithm with OP's requirements, 3 passes, and no external libraries:

public static int Second_Tiny() {

  int[] ar = {19, 1, 17, 17, -2};

  //1st pass - find the smallest item on original array  
  int i;
  int z = ar[0];
  for (i = 1; i < ar.length; i++) {
    if (z >ar[i]){
      z=ar[i];
    }
  }

  //2nd pass copy all items except smallest one to 2nd array
  int[] ar2 = new int[ar.length-1];
  int curIndex = 0;
  for (i=0; i<ar.length; i++) {
    if (ar[i]==z)
      continue;
    ar2[curIndex++] = ar[i];
  }

  //3rd pass - find the smallest item again
  z = ar2[0];
  for (i = 1; i < ar2.length; i++) {
    if (z >ar2[i]){
      z=ar2[i];
    }
  }

  return z;
}

Answer 3

This grabs the index of the element specified in variable z and then sets a second array to the first array minus that one element.

Essentially this gives ar2 = ar1 minus element z

public static int Second_Tiny() {
    int[] ar = {19, 1, 17, 17, -2};
    int[] ar2;
    int i;
    int z = ar[0];
    int x = 0;

    for (i = 1; i < ar.length; i++) {
        if (z >ar[i]){  
            z=ar[i];
            x=i;  
        }                   
    }
    ar2 = ArrayUtils.remove(ar, x);
    return(z);    
}