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Python merge sort in place, so space complexity is O(1)
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import random | |
def merge_sort(xs): | |
"""Inplace merge sort of array without recursive. The basic idea | |
is to avoid the recursive call while using iterative solution. | |
The algorithm first merge chunk of length of 2, then merge chunks | |
of length 4, then 8, 16, .... , until 2^k where 2^k is large than | |
the length of the array | |
""" | |
unit = 1 | |
while unit <= len(xs): | |
h = 0 | |
for h in range(0, len(xs), unit * 2): | |
l, r = h, min(len(xs), h + 2 * unit) | |
mid = h + unit | |
# merge xs[h:h + 2 * unit] | |
p, q = l, mid | |
while p < mid and q < r: | |
# use <= for stable merge merge | |
if xs[p] <= xs[q]: p += 1 | |
else: | |
tmp = xs[q] | |
xs[p + 1: q + 1] = xs[p:q] | |
xs[p] = tmp | |
p, mid, q = p + 1, mid + 1, q + 1 | |
unit *= 2 | |
return xs | |
def test(): | |
assert merge_sort([4,3,2,1]) == [1,2,3,4] | |
assert merge_sort([4,2,3,1]) == [1,2,3,4] | |
assert merge_sort([4,5,3,2,1]) == [1,2,3,4,5] | |
for _ in range(100): | |
tmp = range(100) | |
random.shuffle(tmp) | |
assert merge_sort(tmp) == range(100) | |
return 'test pass!' |
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Thanks for pointing out this. I don't notice about stable sort at first :). Already made the change.