Last night it took me about two hours to learn arrays. For the sake of less time， I did not put emphaises on the practice question, just now when reading the book, I found that some methods referred to arrays are so beneficial to us. So in here make a simple summary.

Method 1: Check whether the array is sorted. 

private static boolean isSorted(int[] a) {    if (a.length < 2) {      return true;    }    for (int i = 1; i < a.length; i++) {      if (a[i] < a[i-1]) {        return false;      }    }    return true;  }}

     Method 2:  Use the start number and range to init the array

public static void load(int[] a, int start, int range) {    for (int i = 0; i < a.length; i++) {      a[i] = start + random.nextInt(range);  // random 5-digit numbers    }  }

    Method 3:  Get the min number from the array

private static int minimum(int[] a) {    int min = a[0];    for (int i = 1; i < a.length; i++) {      if (a[i] < min) {        min = a[i];      }    }    return min;  }

  Method 4： Remove the duplicate elements from object

private static int[] withoutDuplicates(int[] a) {    int n = a.length;    if (n < 2) {      return a;    }    for (int i = 0; i < n-1; i++) {      for (int j = i+1; j < n; j++) {        if (a[j] == a[i]) {          --n;          System.arraycopy(a, j+1, a, j, n-j);          --j;        }      }    }    int[] aa = new int[n];    System.arraycopy(a, 0, aa, 0, n);    return aa;  }

  Method 5: Finds the prime number according to certain range

private static final int SIZE=1000;  private static boolean[] isPrime = new boolean[SIZE];  private static void initializeSieve() {      for (int i = 2; i < SIZE; i++) {        isPrime[i] = true;      }      for (int n = 2; 2*n < SIZE; n++) {        if (isPrime[n]) {          for (int m = n; m*n 

Another way of implement the function of finding the prime number(Vector)

private static final int SIZE=1000;  private static Vector
     
       isPrime = new Vector
      
       (SIZE);    private static void initializeSieve() {    isPrime.add(false);  // 0 is not prime    isPrime.add(false);  // 1 is not prime    for (int i = 2; i < SIZE; i++) {      isPrime.add(true);    }    for (int n = 2; 2*n < SIZE; n++) {      if ((isPrime.get(n))) {        for (int m = n; m*n < SIZE; m++) {          isPrime.set(m*n, false);        }      }    }  }
      
     

Another way of implement the function of finding the prime number(BitSet)

private static final int SIZE=1000;  private static BitSet isPrime = new BitSet(SIZE);  private static void initializeSieve() {      for (int i = 2; i < SIZE; i++) {        isPrime.set(i);      }      for (int n = 2; 2*n < SIZE; n++) {        if (isPrime.get(n)) {          for (int m = n; m*n 

Method 6: Print out the result according to the certain format:

public static void printSieve() {    int n=0;    for (int i = 0; i < SIZE; i++) {      if (isPrime[i]) {        System.out.printf("%5d%s", i, ++n%16==0?"\n":"");      }    }    System.out.printf("%n%d primes less than %d%n", n, SIZE);  }

Notes: There exists five spaces between each number, and it will change line when the length of char  % 6 is zero.

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