【剑指offer】面试题36：数组中的逆序对

# @left part: [start, mid]# @right part: (mid, end]def merge(data, start, mid, end):if mid < start or end < mid:return 0reverse = 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 partst2 = mid + 1while start <= mid and st2 <= end:if data[start] > data[st2]:# move data[st2] to start, and data[start, - 1]# move back by 1data.insert(start, data.pop(st2))reverse += st2 - start# [start, mid] move back by one, so start = start + 1 # and mid = mid + 1start += 1mid += 1# same as start & mid, move back by 1st2 += 1else:# no insert or move, so start = start + 1start += 1return reversedef mergeSortWithCount(data, start, end):#just one item or Node, return dirctlyif len(data) <= 1 or end <= start: return 0mid = (start + end) >> 1#@ reversepair number of left partleft = mergeSortWithCount(data, start, mid)# reversepair number of right partright = 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，还可以通过线段树来实现，有兴趣的可以找一些线段树的题练一练。下面的链接是我学习线段树的第一个博客。

http://www.notonlysuccess.com/index.php/segment-tree/