## Merge Sort in C++This code demonstrates merge sort implemented in C++. Notice the use of a wrapper function to dynamically allocate the requisite scratch space.Back to the merge sort tutorial /* Helper function for finding the max of two numbers */ int max(int x, int y) { if(x > y) { return x; } else { return y; } } /* left is the index of the leftmost element of the subarray; right is one * past the index of the rightmost element */ void merge_helper(int *input, int left, int right, int *scratch) { /* base case: one element */ if(right == left + 1) { return; } else { int i = 0; int length = right - left; int midpoint_distance = length/2; /* l and r are to the positions in the left and right subarrays */ int l = left, r = left + midpoint_distance; /* sort each subarray */ merge_helper(input, left, left + midpoint_distance, scratch); merge_helper(input, left + midpoint_distance, right, scratch); /* merge the arrays together using scratch for temporary storage */ for(i = 0; i < length; i++) { /* Check to see if any elements remain in the left array; if so, * we check if there are any elements left in the right array; if * so, we compare them. Otherwise, we know that the merge must * use take the element from the left array */ if(l < left + midpoint_distance && (r == right || max(input[l], input[r]) == input[l])) { scratch[i] = input[l]; l++; } else { scratch[i] = input[r]; r++; } } /* Copy the sorted subarray back to the input */ for(i = left; i < right; i++) { input[i] = scratch[i - left]; } } } /* mergesort returns true on success. Note that in C++, you could also * replace malloc with new and if memory allocation fails, an exception will * be thrown. If we don't allocate a scratch array here, what happens? * * Elements are sorted in reverse order -- greatest to least */ int mergesort(int *input, int size) { int *scratch = (int *)malloc(size * sizeof(int)); if(scratch != NULL) { merge_helper(input, 0, size, scratch); free(scratch); return 1; } else { return 0; } }Back to the merge sort tutorial |