# @left part: [start, mid]
# @right part: (mid, end]
def merge(data, start, mid, end):
if mid < start or end < mid:
return 0
reverse = 0
'''
@ for start, it play as the start index of left part, and mid
@ play as the end index of left part;
@ mid + 1 (st2) play as the start index of right part, and end
@ play as the end index of right part.
'''
#@ start of right part
st2 = mid + 1
while start <= mid and st2 <= end:
if data[start] > data[st2]:
# move data[st2] to start, and data[start, - 1]
# move back by 1
data.insert(start, data.pop(st2))
reverse += st2 - start
# [start, mid] move back by one, so start = start + 1
# and mid = mid + 1
start += 1
mid += 1
# same as start & mid, move back by 1
st2 += 1
else:
# no insert or move, so start = start + 1
start += 1
return reverse
def mergeSortWithCount(data, start, end):
#just one item or Node, return dirctly
if len(data) <= 1 or end <= start:
return 0
mid = (start + end) >> 1
#@ reversepair number of left part
left = mergeSortWithCount(data, start, mid)
# reversepair number of right part
right = mergeSortWithCount(data, mid + 1, end)
# toal count of reverse pair = left + right + {merger(left, right)}
return merge(data, start, mid, end) + left + right
这个题，还能够通过线段树来实现，有兴趣的能够找一些线段树的题练一练。以下的链接是我学习线段树的第一个博客。