【分治】逆序数对问题：求一个序列中的逆序数对

问题描述

解题思路

代码

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


def sort_and_count(A):
    if len(A) == 1:
        return A, 0
    mid = len(A)//2
    left, count_left = sort_and_count(A[:mid])
    print('left:', left, count_left)
    right, count_right = sort_and_count(A[mid:])
    print('right:', right, count_right)
    sort_A, count_merge = merge_and_count3(left, right)
    print('merge:', sort_A, count_merge)
    return sort_A, count_left + count_right + count_merge


def merge_and_count(left, right):
    result = []
    count_merge = 0
    while len(left) > 0 and len(right) > 0 :
        if left[0] <= right[0]:
            result.append(left.pop(0))
        else:
            count_merge += len(left)
            result.append(right.pop(0))
    result += left
    result += right
    return result, count_merge


# 若题目改成若ai > 3 * aj, 则是逆序，则更改merge_and_cout函数
def merge_and_count3(left, right):
    result = []
    count_merge = 0
    while len(left) > 0 and len(right) > 0:
        if left[0] <= right[0]:
            # 若当前非逆序，还要判断left[0]和right后面的一些
            j = 1
            while j < len(right) and left[0] < 3 * right[j]:  # 直到找到满足条件的为止
                j += 1
            count_merge += (len(right)-j)

            result.append(left.pop(0))
        else:
            if left[0] > 3 * right[0]:  # 若成立，则直接left剩下的全部是逆序
                count_merge += len(left)
            else:  # 若不成立，还得比较
                i = 1
                while i < len(left) and left[i] < 3 * right[0]:  # 直到找到满足条件的为止
                    i += 1
                count_merge += (len(left)-i)

            result.append(right.pop(0))
    result += left
    result += right
    return result, count_merge



A = [3,8,5,7,2,4,9,1]
# A = [2,4,1,3,5]
print(sort_and_count(A))