传说中的华为Python笔试题——两等长整数序列互换元素,序列和的差值最小(修正)

时间:2024-01-03 12:33:02

有两个序列a,b,大小都为n,序列元素的值任意整形数,无序;
要求:通过交换a,b中的元素,使[序列a元素的和]与[序列b元素的和]之间的差最小。

1. 将两序列合并为一个序列,并排序,得到sourceList
2. 拿出最大元素Big,次大的元素Small
3. 在余下的序列S[:-2]进行平分,得到序列max,min
4. 将Small加到max序列,将Big加大min序列,重新计算新序列和,和大的为max,小的为min。

如下,提供递归版本和迭代版本的解体思路:

#!/usr/bin/env python
# -*- coding: utf-8 -*-
#########################################################################
# File Name: minDistance.py
# Author: lpqiu
# mail: qlp_1018@126.com
# Created Time: 2014年10月27日 星期一 05时19分45秒
######################################################################### # recusive version
def minDist(sortedList):
if not sortedList:
return ([], []) large = sortedList[-1]
secondLarge = sortedList[-2] largeList, smallList = minDist(sortedList[:-2])
largeList.append(secondLarge)
smallList.append(large)
#print('LargeList:', largeList, 'SmallList', smallList) if sum(largeList) > sum(smallList):
return (largeList, smallList)
else:
return (smallList, largeList) # iteriation version
def minDistIterVer(sortedList):
largeList, smallList = [], [] for i in range(0, len(sortedList), 2):
if sum(largeList) > sum(smallList):
largeList.append(sortedList[i])
smallList.append(sortedList[i + 1])
else:
largeList.append(sortedList[i + 1])
smallList.append(sortedList[i]) if sum(largeList) > sum(smallList):
return (largeList, smallList)
else:
return (smallList, largeList) def testMain(listA, listB): # len(listA) == len(listB) = n
print("Sorted List: ", sorted(listA + listB))
largeList, smallList = minDist(sorted(listA + listB))
print(largeList, smallList, "distance: ", sum(largeList) - sum(smallList)) largeListIterVer, smallListIterVer = minDistIterVer(sorted(listA + listB))
print(largeListIterVer, smallListIterVer, "distance: ", sum(largeListIterVer) - sum(smallListIterVer)) if __name__ == "__main__":
listA = [1,-1,2,3,6,189]; listB = [2,3,4,6,7,123]
testMain(listA, listB)