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: [email protected] # 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)
时间: 2024-10-01 03:59:47