[英]Find the first longest char occurrence in string lines
int lines = File.ReadAllLines(Path).Length; // count lines index
public string input = "";
public string mtd() // This method should return the first long occurrence index (ry) and the longest occurrence of the char (rx)
{
Dictionary<char, int> dict = new Dictionary<char, int>();
int max = 0;
foreach (char c in input)
{
int i;
dict.TryGetValue(c, out i);
i++;
if (i > max)
{
max = i;
}
dict[c] = i;
}
string rx = "";
string ry = "";
foreach (KeyValuePair<char, int> chars in dict)
{
string x = chars.Key.ToString();
string y = chars.Value.ToString();
if (chars.Value == max)
{
rx = x;
ry = y;
}
}
return rx;
}
我的目标是:
使用一个按钮,使用FileFile.txt(具有更多文本行),使用OpenFileDialog美国。另一个按钮在richTextBox中显示每行的第一个最长子串(char出现)和第一个最长的chars索引(从零开始) 。
这是我的意思的一个例子。
在文本文件中,我有3行:
AAABB
ccddddd
efffggggg
结果应为:
AAA级,0
DDDDD,1
GGGGG,4
我已经看到了有关此问题的其他问题,但没有找到解决方案。 任何想法?
因此,这与其他方法稍有不同,但是您可以尝试使用正则表达式。 以下表达式将匹配重复字符的模式:
(.)(\1+)
(。)匹配换行符以外的任何字符,并且(\\ 1+)匹配先前匹配的1个或多个重复。
如果需要,您可以为第一个匹配组使用(。)以外的名称。 这取决于您关心的重复类型。 例如,如果您只关心重复的“ a”,则可以改用(a)。
这种方法为您提供了很大的灵活性,因为(。)组可以由字符串变量定义,而不必进行硬编码。
算法:
有一个用于正则表达式的系统C#程序集。
https://msdn.microsoft.com/en-us/library/system.text.regularexpressions.regex(v=vs.110).aspx
你可以这样做:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text.RegularExpressions;
namespace ConsoleApplication1
{
public class Program
{
public static void Main()
{
//var lines = File.ReadAllLines("C:\\text.txt");
var lines = new List<string> { "aaabb", "ccddddd", "efffggggg" };
var result = (
from line in lines
let matches = Regex.Matches(line, "(.)\\1+").Cast<Match>()
let maxLen = matches.Max(match => match.Length)
let maxMatch = matches.First(match => match.Length == maxLen)
let index = line.IndexOf(maxMatch.Value)
select string.Format("{0},{1}", maxMatch.Value, index)
).ToList();
result.ForEach(Console.WriteLine);
}
}
}
以下内容将为您提供所需的结果,并在O(n)时间内运行。
var lines = new List<string> { "aaabb", "ccddddd", "efffggggg" };
foreach (var line in lines)
{
if (string.IsNullOrEmpty(line)) // if the line is null or empty then skip it.
{
Console.WriteLine("Empty or Null string.");
continue;
}
char prev = line[0]; // The previous character seen starts with the first character
int maxSeen = 0; // The maximum number of consecutive chars seen
int maxSeenIndex = -1; // The index of the maximum seen chars.
int currentSeen = 1; // The current number of consecutive chars seen.
int currentSeenIndex = 0; // The index of the current chars seen.
for (int i = 1; i < line.Length; i++) // Start at 1 to skip the first character.
{
if (prev == line[i]) // If the current character is the same as the previous
{
currentSeen++; // increment the number of current chars seen.
}
else // If the current character is different
{
if (currentSeen > maxSeen) // Check if the current Seen is more than max
{
maxSeen = currentSeen;
maxSeenIndex = currentSeenIndex;
}
currentSeen = 1; // reset the current seen to 1
currentSeenIndex = i; // set the current seen index to the current index
}
prev = line[i]; // set the current char to the previous
}
if (currentSeen > maxSeen) // Have to do this check again
{
maxSeen = currentSeen;
maxSeenIndex = currentSeenIndex;
}
Console.WriteLine(line.Substring(maxSeenIndex, maxSeen) + ", " + maxSeenIndex);
}
下面是对给定行执行所需操作的函数:
public static string GetCharacterRepetitionAndPosition(string s)
{
if (string.IsNullOrWhiteSpace(s))
return s;
var result = (from ch in s
group ch by ch into g
select new { Cnt = g.Count(), Ch = g.Key });
var maxCnt = -1;
char theMaxChar =char.MinValue;
int howManyCharacters;
foreach (var item in result)
{
if (item.Cnt > maxCnt)
{
maxCnt = item.Cnt;
theMaxChar = item.Ch;
howManyCharacters = item.Cnt;
}
}
var idx = s.IndexOf(theMaxChar);
return new string(theMaxChar,maxCnt) + "," + idx;
}
用法是:
using (var fileStream = new FileStream(Path, FileMode.Open, FileAccess.Read))
{
using (var streamReader = new StreamReader(fileStream))
{
var line = streamReader.ReadLine();
var res = GetCharacterRepetitionAndPosition(line);
// do whatever you want with this output
Console.WriteLine(res);
}
}
算法复杂度分析:
1)groupby的选择为O(N)
2)结果的foreach也为O(N)
3)IndexOf()调用为O(N)
因此总复杂度(大O)为O(N)
这是怎么回事:
1)我们首先按照角色的外貌对所有角色进行分组,然后计算一组中有多少字符。
2)我们对这个结果进行迭代,并且记住角色的最大分配数以及角色是什么
3)我们返回一个字符串(具有最高幻影的字符)
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