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Combining MergeSort with Insertion sort to make it

2020-02-10 10:36发布

问题:

So I have a MergeSort algorithm and I want to combine MergeSort with Insertion sort to reduce the overhead of merging, the question is how? I want to sort the segments using insertion sort and then merge.

 public class mergesorttest{
    public static void main(String[]args){
        int d[]= {10,2,3,4,5,6,5,4,3,5,6,7,1};
        mergeSort(d,0,d.length);
        for(int x:d) System.out.print(x+" "); 
        System.out.println(); 
    }

static void mergeSort(int f[],int lb, int ub){
    //termination reached when a segment of size 1 reached -lb+1=ub
    if(lb+1<ub){
        int mid = (lb+ub)/2;
        mergeSort(f,lb,mid);
        mergeSort(f,mid,ub);
        merge(f,lb,mid,ub);
    }
}

static void merge (int f[],int p, int q, int r){
    //p<=q<=r
    int i =p; int j = q; 
    //use temp array to store merged sub-sequence
    int temp[] = new int[r-p]; int t = 0; 
    while(i<q && j<r){
        if(f[i]<=f[j]){
            temp[t] =f[i]; 
            i++;t++;
        }
        else{
            temp[t] = f[j];
            j++;
            t++;
        }

        //tag on remaining sequence
        while(i<q){
            temp[t] = f[i];
            i++;
            t++;

        }
        while(j<r){
            temp[t]=f[j];
            j++;
            t++;
        }
        //copy temp back to f
        i=p;t=0;
        while(t<temp.length){
            f[i]=temp[t];
            i++;
            t++;
        }
        }
}
}

public static void insertion_srt(int array[], int n, int b){
  for (int i = 1; i < n; i++){
  int j = i;
  int B = array[i];
  while ((j > 0) && (array[j-1] > B)){
  array[j] = array[j-1];
  j--;
  }
  array[j] = B;
  }
  }

回答1:

The merging automatically takes care of sorting the elements. However, one can sort using insertion sort when the list gets below some threshold:

static final int THRESHOLD = 10;
static void mergeSort(int f[],int lb, int ub){
    if (ub - lb <= THRESHOLD)
        insertionSort(f, lb, ub);
    else
    {
        int mid = (lb+ub)/2;
        mergeSort(f,lb,mid);
        mergeSort(f,mid,ub);
        merge(f,lb,mid,ub);
    }
}

Doing anything other than this (except playing around with the threshold a little) will increase the time taken by merge sort.

Although merge sort is O(n log n) and insertion sort is O(n2), insertion sort has better constants and is thus faster on very small arrays. This, this, this and this are a few related questions I found.



回答2:

public static final int K = 5;

public static void insertionSort(int A[], int p, int q) {
    for (int i = p; i < q; i++) {
        int tempVal = A[i + 1];
        int j = i + 1;
        while (j > p && A[j - 1] > tempVal) {
            A[j] = A[j - 1];
            j--;
        }
        A[j] = tempVal;
    }
    int[] temp = Arrays.copyOfRange(A, p, q +1);
    Arrays.stream(temp).forEach(i -> System.out.print(i + " "));
    System.out.println();
}

public static void merge(int A[], int p, int q, int r) {
    int n1 = q - p + 1;
    int n2 = r - q;
    int[] LA = Arrays.copyOfRange(A, p, q +1);
    int[] RA = Arrays.copyOfRange(A, q+1, r +1);
    int RIDX = 0;
    int LIDX = 0;
    for (int i = p; i < r - p + 1; i++) {
        if (RIDX == n2) {
            A[i] = LA[LIDX];
            LIDX++;
        } else if (LIDX == n1) {
            A[i] = RA[RIDX];
            RIDX++;
        } else if (RA[RIDX] > LA[LIDX]) {
            A[i] = LA[LIDX];
            LIDX++;
        } else {
            A[i] = RA[RIDX];
            RIDX++;
        }
    }
}

public static void sort(int A[], int p, int r) {
    if (r - p > K) {
        int q = (p + r) / 2;
        sort(A, p, q);
        sort(A, q + 1, r);
        merge(A, p, q, r);
    } else {
        insertionSort(A, p, r);
    }
}

public static void main(String string[]) {
    int[] A = { 2, 5, 1, 6, 7, 3, 8, 4, 9 };
    sort(A, 0, A.length - 1);
    Arrays.stream(A).forEach(i -> System.out.print(i + " "));
}