Following is a typical recursive implementation of Merge Sort that uses last element as pivot.
C
/* Recursive C program for merge sort */#include <stdio.h> #include <stdlib.h> /* Function to merge the two haves arr[l..m] and arr[m+1..r] of array arr[] */void merge(int arr[], int l, int m, int r); /* l is for left index and r is right index of the sub-array of arr to be sorted */void mergeSort(int arr[], int l, int r) { if (l < r) { int m = l + (r - l) / 2; // Same as (l+r)/2 but avoids overflow for large l & h mergeSort(arr, l, m); mergeSort(arr, m + 1, r); merge(arr, l, m, r); } } /* Function to merge the two haves arr[l..m] and arr[m+1..r] of array arr[] */void merge(int arr[], int l, int m, int r) { int i, j, k; int n1 = m - l + 1; int n2 = r - m; /* create temp arrays */ int L[n1], R[n2]; /* Copy data to temp arrays L[] and R[] */ for (i = 0; i < n1; i++) L[i] = arr[l + i]; for (j = 0; j < n2; j++) R[j] = arr[m + 1 + j]; /* Merge the temp arrays back into arr[l..r]*/ i = 0; j = 0; k = l; while (i < n1 && j < n2) { if (L[i] <= R[j]) { arr[k] = L[i]; i++; } else { arr[k] = R[j]; j++; } k++; } /* Copy the remaining elements of L[], if there are any */ while (i < n1) { arr[k] = L[i]; i++; k++; } /* Copy the remaining elements of R[], if there are any */ while (j < n2) { arr[k] = R[j]; j++; k++; } } /* Function to print an array */void printArray(int A[], int size) { int i; for (i = 0; i < size; i++) printf("%d ", A[i]); printf("\n"); } /* Driver program to test above functions */int main() { int arr[] = { 12, 11, 13, 5, 6, 7 }; int arr_size = sizeof(arr) / sizeof(arr[0]); printf("Given array is \n"); printArray(arr, arr_size); mergeSort(arr, 0, arr_size - 1); printf("\nSorted array is \n"); printArray(arr, arr_size); return 0; } |
Output:
Given array is 12 11 13 5 6 7 Sorted array is 5 6 7 11 12 13
Time Complexity: O(n*log(n))
Auxiliary Space: O(n)
Please refer complete article on Iterative Merge Sort for more details!
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