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 <SIZE; m++) {
isPrime[m*n] = false;
}
}
}
}
Another way of implement the function of finding the prime number(Vector)
private static final int SIZE=1000;
private static Vector<Boolean> isPrime = new Vector<Boolean>(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 <SIZE; m++) {
isPrime.clear(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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