Topcoder上的一道题目，题目描述如下：

Problem Statement

Byteland is a city with many skyscrapers, so it's a perfect venue for BASE jumping. Danilo is an enthusiastic BASE jumper. He plans to come to Byteland and to jump off some of its buildings.

Danilo wants to make M jumps in Byteland. However, he has some rules. First, he never jumps off the same building twice. Second, all buildings he selects for his jumps must have the same number of floors. (This is for safety reasons: It is hard to get the timing right if each jump starts at a different height.)

Philipe is the mayor of Byteland. He welcomes Danilo's visit as he would like to use it as a publicity stunt. However, Philipe knows that Danilo will only come to Byteland if there are at least M buildings that each have the same number of floors. To ensure that, the mayor is willing to build additional floors on some of the skyscrapers in Byteland.

You are given the int M and a int[] heights. Each element of heights is the number of floors in one of Byteland's skyscrapers. Compute and return the smallest number of additional floors the mayor has to build so that there will be at least M buildings with the same number of floors.

Definition

Class:

BuildingHeightsEasy

Method:

minimum

Parameters:

int, int[]

Returns:

int

Method signature:

int minimum(int M, int[] heights)

(be sure your method is public)

Limits

Time limit (s):

2.000

Memory limit (MB):

256

Constraints

-

heights will contain between 1 and 50 elements, inclusive.

-

M will be between 1 and the number of elements in heights, inclusive.

-

Each element in heights will be between 1 and 50, inclusive.

Examples

0)

2

{1, 2, 1, 4, 3}

Returns: 0

Note that we already have two buildings that have the same number of floors. Hence, no additional floors need to be built.

1)

3

{1, 3, 5, 2, 1}

Returns: 2

We want to have at least three buildings with the same number of floors. The best way to reach this goal is to build one floor on building #0 and one floor on building #4 (0-based indices). After these changes, buildings #0, #3, and #4 will have two floors each.

2)

1

{43, 19, 15}

Returns: 0

3)

3

{19, 23, 9, 12}

Returns: 15

4)

12

{25, 18, 38, 1, 42, 41, 14, 16, 19, 46, 42, 39, 38, 31, 43, 37, 26, 41, 33, 37, 45, 27, 19, 24, 33, 11, 22, 20, 36, 4, 4}

Returns: 47

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View Code

题目意思是给定一堆楼高度的 List 为 heights ，需要在这些楼里选择 M 栋高度相同的楼，如果没够足够相同高度的楼，就加高一些楼层使得满足需求。求至少需要加高多少层数楼。

最直接的解题思路：

排序所有楼层，分别计算加高到某一楼层需要的总层数，取出其中最少的数字。

假设所有楼层有：{19, 23, 9, 12} 4栋楼。要在其中取 M= 3 栋相同的楼层，排序后的序列为：

{9, 12, 19, 23}

则从第 3 栋开始考虑增加楼层，即将 9， 12 增加至 19 ， 共需增加 19*3 - (9 + 12 + 19) = 17 层

在第 4 栋楼层(23)开始增加：共需增加 23*3 - (12 + 19 + 23) = 15 层

答案是 15

Java 写的实现：

importjava.util.ArrayList;importjava.util.Collections;importjava.util.List;public classBuildingHeightsEasy {public static voidmain(String[] args) {

BuildingHeightsEasy bh= newBuildingHeightsEasy();int M = 3;int[] heights = {19, 23, 9, 12} ;int total =bh.minimum(M, heights);

System.out.println(total);

}public int minimum(int M, int[] heights) {if (M == 1) {return 0;

}

List list = new ArrayList();for(Integer cost : heights) {

list.add(cost);

}

Collections.sort(list);

Integer total=Integer.MAX_VALUE;for (int i = M - 1; i < list.size(); i++) {int sum =getSumValue(M, i, list);if (sum

total=sum;

}

}returntotal;

}public int getSumValue(int M, int end, Listlist) {int sum = 0;for (int i = end - M + 1; i < end + 1; i++) {

sum+=list.get(i);

}int value =list.get(end);return value * M -sum;

}

}

Java 在 int 数组转换为 List 和 sum List 值等方便会比较繁琐。

Python 的实现：

classBuildingHeightsEasy(object):defminimum(self, M, heights):if M == 1 : return0

heights=list(heights)

heights.sort()

total= M * heights[len(heights)-1]for i in range(M-1,len(heights)):

sumValue= M * heights[i] - sum(heights[i-M+1:i+1])

total= sumValue if sumValue < total elsetotalreturntotal

M= 3heights= (19, 23, 9, 12)

bh=BuildingHeightsEasy()print bh.minimum(M,heights)

Python 内置了很丰富的函数，特别是 sum 函数和 list 的切片功能。

结语留空。