Merge sorting is a good known algorithm, but usually is used recursive implementation with additional array. This is an another version: without recursive and without usage of additional array.
Merge sorting idea is to sort small parts (pairs) than longer (4 items) and etc, but each next sort phase is MERGE of previous sorted parts. Merging of 2 sorted parts should saves order of items:
1,5 merging 2,3 is 1,2,3,5.
If we avoid recursive implementation then we should generate indexes of all parts (2 items, 4 items...). This sub-task will be done in mergesort() function. Another sub-task is to merge 2 sorted parts into new one, but sorted too. This is implemented in _merge() function, which avoid additional array of result by using of swapping of items: we has 2 arrays with 2 indexes and compare both items (from one array and another, indexed by these 2 indexes). If they need reordering then we swap items. Index of lesser item is incremented (see code below).
Main problem is to calculate middle index (for splitting input array into 2 parts for merging). Next formula is used:
mid = lev*2 + pielen*pie;
if (pielen > len || mid > len-1) mid = lastleft;
where
lev is the level of pseudo-recursion,
pielen is the length of pie (part) = {2, 4, 8...}.
len is the length of input array,
lastleft is the index of left bound of previous "level". So, we should to get something like this:
lev=0 left=0 mid=0 right=2
lev=0 left=2 mid=2 right=4
lev=0 left=4 mid=4 right=6
lev=0 left=6 mid=6 right=8
lev=0 left=8 mid=8 right=10
lev=0 left=10 mid=10 right=12
lev=1 left=0 mid=2 right=4
lev=1 left=4 mid=6 right=8
lev=1 left=8 mid=10 right=12
lev=2 left=0 mid=4 right=8
lev=2 left=8 mid=8 right=12
lev=3 left=0 mid=8 right=12
So, this is the code:
#include <stdio.h>
#include "algutils.h"
template<typename>
static inline void _merge(T *arr, int left, int mid, int right)
{
T tmp, // tmp item needed for inserting if head position on shift
*larr, // left subarray
*rarr; // right subarray
int llen, // length of left subarray
rlen; // length of right subarray
int l, r; // indexes of compared items in left, right subarrays
l = r = 0;
larr = &arr[left];
if (mid==left) {
// When len==2 swapping is more preferable
rarr = &arr[mid+1];
llen = mid - left + 1;
rlen = right - mid - 1;
} else {
// Common case
rarr = &arr[mid];
llen = mid - left;
rlen = right - mid;
}
// Index in left subarray is limited by "right" index
while ((larr+l < rarr+r) && r < rlen) {
if (larr[l] > rarr[r]) {
tmp = rarr[r];
RSHIFT_ARR(larr+l, llen-(l-r), 1);
larr[l] = tmp;
r++;
}
l++;
}
}
template<typename>
int mergsort(T* arr, int len)
{
int lev, // level of "recursion"
pie, // number of subarray (of part)
pielen, // subarray length: 2, 4, 8...
left, right, // bounds (right is not included)
mid, // middle point for splitting - is the same as left bound of subarray but on prev. level
lastleft; // last left index
if (!arr||!len) return (0);
lev = 0;
pielen = 2;
while (pielen < 2*len) {
pie = 0;
do {
left = pie*pielen;
right = left + pielen;
if (right > len) right = len;
mid = lev*2 + pielen*pie;
if (pielen > len || mid > len-1) mid = lastleft; //mid = len-2;
_merge(arr, left, mid, right);
pie++;
} while (right < len-1);
pielen *= 2;
lev++;
lastleft = left;
}
return (1);
}
/*****************************************
main
****************************************/
int main(void)
{
int arr1[] = {10, 19, 1, 2, 10, 450, 6, 7, 8, 100, 120, 121};
mergsort(arr1, sizeofarray(arr1));
PRINTARR(arr1, "%d");
return (0);
}
Macros RSHIFT_ARR() is trivial:
/// Shift array @a ARR of length @a LEN by @a N positions to right
#define RSHIFT_ARR(ARR, LEN, N) do { \
register int _rshift_arr_i_; \
for (_rshift_arr_i_=(N)+(LEN)-1; \
_rshift_arr_i_>(N)-1; \
_rshift_arr_i_--) \
{ \
*((ARR)+_rshift_arr_i_) = *((ARR)+(_rshift_arr_i_-(N))); \
} \
} while (0)
