algorithm-in-python/sort/heapSort.py

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2018-07-08 23:28:29 +08:00
''' mbinary
#########################################################################
# File : heapSort.py
# Author: mbinary
# Mail: zhuheqin1@gmail.com
2019-01-31 12:09:46 +08:00
# Blog: https://mbinary.xyz
2018-07-08 23:28:29 +08:00
# Github: https://github.com/mbinary
# Created Time: 2018-07-05 16:24
# Description:
#########################################################################
'''
2018-07-08 16:41:55 +08:00
from functools import partial
class heap:
def __init__(self,lst,reverse = False):
self.data= heapify(lst,reverse)
self.cmp = partial(lambda i,j,r:cmp(self.data[i],self.data[j],r),r= reverse)
def getTop(self):
return self.data[0]
def __getitem__(self,idx):
return self.data[idx]
def __bool__(self):
return self.data != []
def popTop(self):
ret = self.data[0]
n = len(self.data)
cur = 1
while cur * 2<=n:
chd = cur-1
r_idx = cur*2
l_idx = r_idx-1
if r_idx==n:
self.data[chd] = self.data[l_idx]
break
j = l_idx if self.cmp(l_idx,r_idx)<0 else r_idx
self.data[chd] = self.data[j]
cur = j+1
self.data[cur-1] = self.data[-1]
self.data.pop()
return ret
def addNode(self,val):
self.data.append(val)
self.data = one_heapify(len(self.data)-1)
def cmp(n1,n2,reverse=False):
fac = -1 if reverse else 1
if n1 < n2: return -fac
elif n1 > n2: return fac
return 0
def heapify(lst,reverse = False):
for i in range(len(lst)):
lst = one_heapify(lst,i,reverse)
return lst
def one_heapify(lst,cur,reverse = False):
cur +=1
while cur>1:
chd = cur-1
prt = cur//2-1
if cmp(lst[prt],lst[chd],reverse)<0:
break
lst[prt],lst[chd] = lst[chd], lst[prt]
cur = prt+1
return lst
def heapSort(lst,reverse = False):
lst = lst.copy()
hp = heap(lst,reverse)
ret = []
while hp:
ret.append(hp.popTop())
return ret
if __name__ == '__main__':
from random import randint
n = randint(10,20)
lst = [randint(0,100) for i in range(n)]
print('random : ', lst)
print('small-heap: ', heapify(lst))
print('big-heap : ', heapify(lst,True))
print('ascend : ', heapSort(lst))
print('descend : ', heapSort(lst,True))