How to merge two sorted arrays into a sorted array? [closed]

Answer 1

My favorite programming language is JavaScript

function mergeSortedArrays(a, b){
    var result = [];

    var sI = 0;
    var lI = 0;
    var smallArr;
    var largeArr;
    var temp;

    if(typeof b[0] === 'undefined' || a[0]<b[0]){
        smallArr = a;
        largeArr = b;
    } else{
        smallArr = b;
        largeArr = a;
    }

    while(typeof smallArr[sI] !== 'undefined'){
        result.push(smallArr[sI]);
        sI++;

        if(smallArr[sI]>largeArr[lI] || typeof smallArr[sI] === 'undefined'){
            temp = smallArr;
            smallArr = largeArr;
            largeArr = temp;
            temp = sI;
            sI = lI;
            lI = temp;
        }
    }
    return result;
}

Answer 2

You can use ternary operators for making the code a bit more compact

public static int[] mergeArrays(int[] a1, int[] a2) {
    int[] res = new int[a1.length + a2.length];
    int i = 0, j = 0;

    while (i < a1.length && j < a2.length) {
        res[i + j] = a1[i] < a2[j] ? a1[i++] : a2[j++];
    }

    while (i < a1.length) {
        res[i + j] = a1[i++];
    }

    while (j < a2.length) {
        res[i + j] = a2[j++];
    }

    return res;
}

Answer 3

    public class Merge {

    // stably merge a[lo .. mid] with a[mid+1 .. hi] using aux[lo .. hi]
    public static void merge(Comparable[] a, Comparable[] aux, int lo, int mid, int hi) {

        // precondition: a[lo .. mid] and a[mid+1 .. hi] are sorted subarrays
        assert isSorted(a, lo, mid);
        assert isSorted(a, mid+1, hi);

        // copy to aux[]
        for (int k = lo; k <= hi; k++) {
            aux[k] = a[k]; 
        }

        // merge back to a[]
        int i = lo, j = mid+1;
        for (int k = lo; k <= hi; k++) {
            if      (i > mid)              a[k] = aux[j++];
            else if (j > hi)               a[k] = aux[i++];
            else if (less(aux[j], aux[i])) a[k] = aux[j++];
            else                           a[k] = aux[i++];
        }

        // postcondition: a[lo .. hi] is sorted
        assert isSorted(a, lo, hi);
    }

    // mergesort a[lo..hi] using auxiliary array aux[lo..hi]
    private static void sort(Comparable[] a, Comparable[] aux, int lo, int hi) {
        if (hi <= lo) return;
        int mid = lo + (hi - lo) / 2;
        sort(a, aux, lo, mid);
        sort(a, aux, mid + 1, hi);
        merge(a, aux, lo, mid, hi);
    }

    public static void sort(Comparable[] a) {
        Comparable[] aux = new Comparable[a.length];
        sort(a, aux, 0, a.length-1);
        assert isSorted(a);
    }


   /***********************************************************************
    *  Helper sorting functions
    ***********************************************************************/

    // is v < w ?
    private static boolean less(Comparable v, Comparable w) {
        return (v.compareTo(w) < 0);
    }

    // exchange a[i] and a[j]
    private static void exch(Object[] a, int i, int j) {
        Object swap = a[i];
        a[i] = a[j];
        a[j] = swap;
    }


   /***********************************************************************
    *  Check if array is sorted - useful for debugging
    ***********************************************************************/
    private static boolean isSorted(Comparable[] a) {
        return isSorted(a, 0, a.length - 1);
    }

    private static boolean isSorted(Comparable[] a, int lo, int hi) {
        for (int i = lo + 1; i <= hi; i++)
            if (less(a[i], a[i-1])) return false;
        return true;
    }


   /***********************************************************************
    *  Index mergesort
    ***********************************************************************/
    // stably merge a[lo .. mid] with a[mid+1 .. hi] using aux[lo .. hi]
    private static void merge(Comparable[] a, int[] index, int[] aux, int lo, int mid, int hi) {

        // copy to aux[]
        for (int k = lo; k <= hi; k++) {
            aux[k] = index[k]; 
        }

        // merge back to a[]
        int i = lo, j = mid+1;
        for (int k = lo; k <= hi; k++) {
            if      (i > mid)                    index[k] = aux[j++];
            else if (j > hi)                     index[k] = aux[i++];
            else if (less(a[aux[j]], a[aux[i]])) index[k] = aux[j++];
            else                                 index[k] = aux[i++];
        }
    }

    // return a permutation that gives the elements in a[] in ascending order
    // do not change the original array a[]
    public static int[] indexSort(Comparable[] a) {
        int N = a.length;
        int[] index = new int[N];
        for (int i = 0; i < N; i++)
            index[i] = i;

        int[] aux = new int[N];
        sort(a, index, aux, 0, N-1);
        return index;
    }

    // mergesort a[lo..hi] using auxiliary array aux[lo..hi]
    private static void sort(Comparable[] a, int[] index, int[] aux, int lo, int hi) {
        if (hi <= lo) return;
        int mid = lo + (hi - lo) / 2;
        sort(a, index, aux, lo, mid);
        sort(a, index, aux, mid + 1, hi);
        merge(a, index, aux, lo, mid, hi);
    }

    // print array to standard output
    private static void show(Comparable[] a) {
        for (int i = 0; i < a.length; i++) {
            StdOut.println(a[i]);
        }
    }

    // Read strings from standard input, sort them, and print.
    public static void main(String[] args) {
        String[] a = StdIn.readStrings();
        Merge.sort(a);
        show(a);
    }
}

Answer 4

I think introducing the skip list for the larger sorted array can reduce the number of comparisons and can speed up the process of copying into the third array. This can be good if the array is too huge.

Answer 5

public static int[] merge(int[] a, int[] b) {

    int[] answer = new int[a.length + b.length];
    int i = 0, j = 0, k = 0;

    while (i < a.length && j < b.length)  
       answer[k++] = a[i] < b[j] ? a[i++] :  b[j++];

    while (i < a.length)  
        answer[k++] = a[i++];

    while (j < b.length)    
        answer[k++] = b[j++];

    return answer;
}

Is a little bit more compact but exactly the same!

Answer 6

Here is updated function. It removes duplicates, hopefully someone will find this usable:

public static long[] merge2SortedAndRemoveDublicates(long[] a, long[] b) {
    long[] answer = new long[a.length + b.length];
    int i = 0, j = 0, k = 0;
    long tmp;
    while (i < a.length && j < b.length) {
        tmp = a[i] < b[j] ? a[i++] : b[j++];
        for ( ; i < a.length && a[i] == tmp; i++);
        for ( ; j < b.length && b[j] == tmp; j++);
        answer[k++] = tmp;
    }
    while (i < a.length) {
        tmp = a[i++];
        for ( ; i < a.length && a[i] == tmp; i++);
        answer[k++] = tmp;
    }
    while (j < b.length) {
        tmp = b[j++];
        for ( ; j < b.length && b[j] == tmp; j++);
        answer[k++] = tmp;
    }
    return Arrays.copyOf(answer, k);
}