Question

这是我在mergesort算法中找到计数反转的程序，虽然我无法找到正确的反转值。谁能告诉我哪里出错了？

import time
import random
start=time.time()

def merge_sort(items):
    """ Implementation of mergesort """
    count=0
    if len(items) > 1:

        mid = len(items) // 2        # Determine the midpoint and split
        left = items[0:mid]
        right = items[mid:]

        merge_sort(left)            # Sort left list in-place
        merge_sort(right)           # Sort right list in-place

        l, r = 0, 0
        for i in range(len(items)):     # Merging the left and right list

            lval = left[l] if l < len(left) else None
            rval = right[r] if r < len(right) else None

            if (lval is not None and rval is not None and lval < rval) or rval is None:
                items[i] = lval
                l += 1
            elif (lval is not None and rval is not None and lval >= rval) or lval is None:
                items[i] = rval
                count=count+(len(items)-l)
                r += 1

        print(count)        
    return items

#n=int(input("enter number of elements in list"))
items=[2,1,5,4,3,9]
#for x in range (n):

 #   items.append(random.random())


print(merge_sort(items))
print("time is %s"%(time.time()-start))

Answer 1

这是一个计算数字列表中的反转次数的函数。

def count_inversions(items):
    inversions = 0
    for idx, item in enumerate(items):
        if idx < len(items) - 1 and items[idx] > items[idx + 1]:
            inversions += 1
    return inversions

你需要打电话一次。它高于任何递归调用（左和右）。

如何在Python中使用mergesort查找计数反转？

1 个答案: