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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【DataStructure】Some useful methods for arrays