将列表转换为数字范围字符串

Question

这个问题几乎与这个问题相反: C#是否内置支持解析页码字符串？

所以给定

1,3,5,6,7,8,9,10,12:

我会输出:

1,3,5-10,12

这是我的第一次尝试.它看起来有点像hacky,可能是我写过的最糟糕的代码.你能提出一个改进的方法吗？更好的办法吗？

static string numListToRangeStr(List<int> numList)
{
    StringBuilder retString = new StringBuilder();
    numList.Sort();

    bool inRangeFind = false;
    int firstInRange = numList[0];
    int lastNumber = firstInRange;
    bool first = true;

    for (int i = 1; i < numList.Count; i++)
    {
        if (numList[i] == (lastNumber + 1))
        {
            inRangeFind = true;
        }
        else
        {             
            if (inRangeFind)
            {
                if (!first)
                {
                    retString.Append(",");
                }
                retString.Append(firstInRange);
                retString.Append("-");
            }
            else
            {
               if (!first)
                {
                    retString.Append(",");
                }
            }

            retString.Append(lastNumber);

            firstInRange = numList[i];
            inRangeFind = false;
            first = false;
        }

        lastNumber = numList[i];
    }


    if (inRangeFind)
    {
        if (!first)
        {
            retString.Append(",");
        }
        retString.Append(firstInRange);
        retString.Append("-");
    }
    else
    {
        if (!first)
        {
            retString.Append(",");
        }
    }
    retString.Append(lastNumber);

    return retString.ToString();
}

Answer 1

当某些东西有这样的几个运动部件时,我认为将它分解成很小的逻辑单元然后将它们组合在一起是有帮助的.小逻辑单元甚至可以单独使用.下面的代码将问题分解为:

• 将异构的顺序和非顺序数集转换为一组同质的范围(可能包括以相同数字开始和结束的"简并"范围)
• 一种"漂亮印刷"这种范围的方法:(x,y)打印为"xy"; (x,x)打印为"x"
• 一种在可枚举元素之间插入分隔符并将结果转换为字符串的方法.

该计划是:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication37 {
  public static class Program {
    static void Main(string[] args) {
      var numList=new[] {1, 3, 5, 6, 7, 8, 9, 10, 12};
      Console.WriteLine(numListToPossiblyDegenerateRanges(numList).Select(r => PrettyRange(r)).Intersperse(","));
    }

    /// <summary>
    /// e.g. 1,3,5,6,7,8,9,10,12
    /// becomes
    /// (1,1),(3,3),(5,10),(12,12)
    /// </summary>
    public static IEnumerable<Tuple<int,int>> numListToPossiblyDegenerateRanges(IEnumerable<int> numList) {
      Tuple<int, int> currentRange=null;
      foreach(var num in numList) {
        if(currentRange==null) {
          currentRange=Tuple.Create(num, num);
        } else if(currentRange.Item2==num-1) {
          currentRange=Tuple.Create(currentRange.Item1, num);
        } else {
          yield return currentRange;
          currentRange=Tuple.Create(num, num);
        }
      }
      if(currentRange!=null) {
        yield return currentRange;
      }
    }

    /// <summary>
    /// e.g. (1,1) becomes "1"
    /// (1,3) becomes "1-3"
    /// </summary>
    /// <param name="range"></param>
    /// <returns></returns>
    public static string PrettyRange(Tuple<int,int> range) {
      if(range.Item1==range.Item2) {
        return range.Item1.ToString();
      }
      return string.Format("{0}-{1}", range.Item1, range.Item2);
    }

    public static string Intersperse(this IEnumerable<string> items, string interspersand) {
      var currentInterspersand="";
      var result=new StringBuilder();
      foreach(var item in items) {
        result.Append(currentInterspersand);
        result.Append(item);
        currentInterspersand=interspersand;
      }
      return result.ToString();
    }
  }
}

Answer 2

这是一个老线程,但这是一个新的答案.我把它构建为扩展方法.这将返回范围数组,其中每个"范围"是单个数字('13')或一对数字('5-12'):

public static class EnumExt {
    public static string[] ToRanges(this List<int> ints) {
        if (ints.Count < 1) return new string[] { };
        ints.Sort();
        var lng = ints.Count;
        var fromnums = new List<int>();
        var tonums = new List<int>();
        for (var i = 0; i < lng - 1; i++) {
            if (i == 0)
                fromnums.Add(ints[0]);
            if (ints[i + 1] > ints[i] + 1) {
                tonums.Add(ints[i]);
                fromnums.Add(ints[i + 1]);
            }
        }
        tonums.Add(ints[lng - 1]);
        return Enumerable.Range(0, tonums.Count).Select(
            i => fromnums[i].ToString() +
                (tonums[i] == fromnums[i] ? "" : "-" + tonums[i].ToString())
        ).ToArray();
    }
}

如果您想加入它们,只需使用内置string.Join:

var intlist = new List<int>() { 1, 2, 3, 6, 7, 8, 9, 10, 14 };
Console.WriteLine(string.Join(", ", intlist.ToRanges()));
// Prints: 1-3, 6-10, 14

Answer 3

必须解决同样的问题。正在寻找我认为看起来更合乎逻辑的解决方案的替代方案。因此分享它。如果要对未排序的列表进行排序，请将第二个参数设置为 true。

public string ToRangeString(List<int> list, bool withSort) {
  list = list.Distinct().ToList();
  if(withSort) list.Sort();

  StringBuilder result = new StringBuilder();
  int temp;

  for (int i=0; i<list.Count(); i++) {
    temp = list[i];

    //add a number
    result.Append(list[i]);

    //skip number(s) between a range
    while(i<list.Count()-1 && list[i+1] == list[i]+1)
      i++;

    //add the range
    if(temp != list[i])
      result.Append("-").Append(list[i]);

    //add comma
    if(i != list.Count()-1)
      result.Append(", ");

  }
  return result.ToString();
}