算法导论习题2-4 Inversions 逆序数

简单地改一下归并排序即可

# -*- coding: utf-8 -*-

def inversioncountmask(l):
l1 = l[:]
return inversionscount(l1, 0, len(l1)-1)

def inversionscount(l, p, r):
if p < r:
q = (p+r)/2
return inversionscount(l, p, q) + inversionscount(l, q+1, r) + mergecount(l, p, q, r)
else:
return 0

def mergecount(l, p, q, r):
l1 = l[p:q+1]
l2 = l[q+1:r+1]
count = 0
for x in range(p, r+1):
if (len(l1) is not 0) & (len(l2) is not 0):
if l1[0] < l2[0]:
l[x] = l1[0]
l1.pop(0)
else:
l[x] = l2[0]
l2.pop(0)
count += len(l1)
elif len(l1) is 0:
l[x] = l2[0]
l2.pop(0)
else:
l[x] = l1[0]
l1.pop(0)
return count

l = [1, 3, 5, 7, 9, 11]
l1 = [2, 3, 8, 6, 1]
print inversioncountmask(l)
print inversioncountmask(l1)