Monday, December 5, 2011

Data Structure: How to implement Merge Sort in C++?

Merging is the process of combining two or more sorted files into a third sorted file. Merge sort is based on the divide and conquer paradigm. It can be explained as …. Let A[p…..r] is given array with indices p = 1 to r = n. Then steps can be

          Image Source: http://en.wikipedia.org/wiki/Merge_sort 
  1. Divide Step: If a given array A has zero or one element, then simply return; it is already sorted. Otherwise split A[p…..r] into two subarrays as A[p….q] and A[q+1…..r] each containing about the half of the elements.
  2. Conquer Step: Conquer by recursively sorted the two sub-arrays A[p….q] and A[q+1….r]
  3. Combine Step: Combine the elements back in A[p…..r] by merging the two sorted sub-arrays into the sorted sequence. Note that: the recursion bottoms out when the sub-array has just one element. So that it is trivially sorted.
Source Code:

#include<iostream>
using namespace std;

//create a class MergeSort
class MergeSort{
    public:
        int no_of_elements;
        int elements[10];
    public:
        void getarray();
        void partition(int [] ,int ,int);
        void sortit(int [], int , int, int);
        void display();
};

// get the array to be sorted from the user
void MergeSort::getarray(){
    cout<<"How many elements?: ";
    cin>>no_of_elements;
    cout<<"Insert elements to sort: ";
    for(int i=0;i<no_of_elements;i++){
        cin>>elements[i];
    }
}

// recursively divide the array into sub arrays until all sub
// arrays contain only one element
void MergeSort::partition(int elements[], int low, int high){
    int mid;
    if(low<high){
        mid=(low+high)/2;
        // sort the left sub array
        partition(elements,low,mid);

        // sort the right sub array
        partition(elements,mid+1,high);


        sortit(elements,low,mid,high);

    }
}
void MergeSort::sortit(int elements[], int low, int mid, int high){
    int i,j,k,l,b[20];
    l=low;
    i=low;
    j=mid+1;
    while((l<=mid)&&(j<=high)){
        if(elements[l]<=elements[j]){
            b[i]=elements[l];
            l++;
        }else{
            b[i]=elements[j];
            j++;
        }
        i++;
    }
    if(l>mid){
        for(k=j;k<=high;k++){
            b[i]=elements[k];
            i++;
        }
    }else{
        for(k=l;k<=mid;k++){
            b[i]=elements[k];
            i++;
        }
    }
    for(k=low;k<=high;k++){
        elements[k]=b[k];
    }
}
void MergeSort::display(){
    cout<<"The sorted element is\n";
    for(int i = 0 ; i < no_of_elements; i++){
        cout<<elements[i]<<" ";
    }
}
int main(){
    MergeSort MS;
    MS.getarray();
    MS.partition(MS.elements,0,MS.no_of_elements);
    MS.display();
    return 0;
}
Output:
How many elements? 7
Insert element to sort: 63 78 52 12 99 32 13
The sorted element is
12 13 32 52 63 78 99

Efficiency of merge sort
Since the merge sort can not have more than logn passes and each pass involving ‘n’ or fewer comparisons, the merge sort requires no more than nlogn comparisons.  Hence the efficiency of merge sort is O(nlogn).

1 comment:

  1. very helpfull. may your companies touch sky. :-)

    ReplyDelete