Question

这是一个问题，你有一堆圆柱体，你移除顶部圆柱体，直到堆叠高度相等。

示例输入：

堆叠有5个，3个和4个圆柱体，这些圆柱体的相应高度在下一行给出。

预期产出：

我的节目输出：

我的节目：

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
class Solution
{
    static void Main(String[] args) 
    {
        Func<string,int[]> GetIntsFromLine = (line) => Array.ConvertAll(line.Split(' '), Int32.Parse);
        int numcyls = GetIntsFromLine(Console.ReadLine()).Length;
        int[][] cyls = new int[numcyls][]; 
        for(int i = 0; i < numcyls; ++i)
            cyls[i] = GetIntsFromLine(Console.ReadLine());
        // now the j-th cylinder stacked on the i-th stach has height cyls[i][j]

        while(true)
        {
              var heights = from cyl in cyls
                            select cyl.Sum();
              int h1 = heights.First();
              if(heights.All(h => h == h1)) // if equal heights
              {
                  Console.WriteLine(h1);
                  return; 
              }
              // if here, "remove" the top cylinder from the stack with the largest height
              int[] st = cyls[heights.ToList().IndexOf(heights.Max())];
              int j = Array.IndexOf(st, 0);
              st[j == -1 ? (st.Length - 1) : j] = 0;            
        }

    }
}

Answer 1

好的，让我们解决问题。所以我们有三个堆栈：

1, 1, 1, 2, 3
2, 3, 4
1, 4, 1, 1

取0,1,2,3 ......（等于删除5,4,3,2 ......）项目从我们可以拥有的第一个堆栈

0                 == 0
1                 == 1
1 + 1             == 2
1 + 1 + 1         == 3
1 + 1 + 1 + 2     == 5 <- this is the solution
1 + 1 + 1 + 2 + 3 == 8

子堆。所以我们可以生产

0, 1, 2, 3, 5, 8 (from the 1st)
0, 2, 5, 9       (from the 2nd)
0, 1, 5, 6, 7    (from the 3d)

我们应该采用所有行中的最大值（可能的子包） - 5.实施：

String input =
  "5 3 4\n" +
  "3 2 1 1 1\n" +
  "4 3 2\n" +
  "1 1 4 1";

var raw = input.Split(new char[] { '\r', '\n' }, StringSplitOptions.RemoveEmptyEntries)
  .Skip(1) // We don't need 1st control line
  .Select(line => line
    .Split(new char[] { ' ' }, StringSplitOptions.RemoveEmptyEntries)
    .Reverse() // <- do not forget this!
    .Select(x => int.Parse(x))); 

// All possible substacks per each stack
List<HashSet<int>> data = new List<HashSet<int>>();

foreach (var record in raw) {
  HashSet<int> hs = new HashSet<int>() { 0 };

  data.Add(hs);

  int agg = 0;

  foreach (var item in record)
    hs.Add(agg += item);
}

// Values that are in all substacks:
HashSet<int> intersect = new HashSet<int>(data[0]);

foreach (var line in data)
  intersect.IntersectWith(line); 

// And, finally, the maximum one:
Console.Write(intersect.Max());

我的等堆栈算法中的缺陷在哪里？

1 个答案: