这是一个函数,可以将单词分割成你想要的段数。然后,您可以遍历该列表以将每个段设置为 button.Text。
public List<string> SplitInSegments(string word, int segments)
{
int wordLength = word.Length;
// The remainder tells us how many segments will get an extra letter
int remainder = wordLength % segments;
// The base length of a segment
// This is a floor division, because we're dividing ints.
// So 5 / 3 = 1
int segmentLength = wordLength / segments;
var result = new List<string>();
int startIndex = 0;
for (int i = 0; i < segments; i++)
{
// This segment may get an extra letter, if its index is smaller then the remainder
int currentSegmentLength = segmentLength + (i < remainder ? 1 : 0);
string currentSegment = word.Substring(startIndex, currentSegmentLength);
// Set the startindex for the next segment.
startIndex += currentSegmentLength;
result.Add(currentSegment);
}
return result;
}
用法:
// returns ["ma", "yb", "e"]
var segments = SplitInSegments("maybe", 3);
编辑
我喜欢这是为了教孩子。所以来了。
关于根据特定字母序列拆分字符串的问题:使用正则表达式拆分字符串后,您将拥有一个字符串数组。然后确定拆分字符串中的项目数量,并根据段数进一步连接或拆分:
// sequences to split on first
static readonly string[] splitSequences = {
"el",
"ol",
"bo"
};
static readonly string regexDelimiters = string.Join('|', splitSequences.Select(s => "(" + s + ")"));
// Method to split on sequences
public static List<string> SplitOnSequences(string word)
{
return Regex.Split(word, regexDelimiters).Where(s => !string.IsNullOrEmpty(s)).ToList();
}
public static List<string> SplitInSegments(string word, int segments)
{
int wordLength = word.Length;
// The remainder tells us how many segments will get an extra letter
int remainder = wordLength % segments;
// The base length of a segment
// This is a floor division, because we're dividing ints.
// So 5 / 3 = 1
int segmentLength = wordLength / segments;
var result = new List<string>();
int startIndex = 0;
for (int i = 0; i < segments; i++)
{
// This segment may get an extra letter, if its index is smaller then the remainder
int currentSegmentLength = segmentLength + (i < remainder ? 1 : 0);
string currentSegment = word.Substring(startIndex, currentSegmentLength);
// Set the startindex for the next segment.
startIndex += currentSegmentLength;
result.Add(currentSegment);
}
return result;
}
// Splitword will now always return 3 segments
public static List<string> SplitWord(string word)
{
if (word == null)
{
throw new ArgumentNullException(nameof(word));
}
if (word.Length < 3)
{
throw new ArgumentException("Word must be at least 3 characters long", nameof(word));
}
var splitted = SplitOnSequences(word);
var result = new List<string>();
if (splitted.Count == 1)
{
// If the result is not splitted, just split it evenly.
result = SplitInSegments(word, 3);
}
else if (splitted.Count == 2)
{
// If we've got 2 segments, split the shortest segment again.
if (splitted[1].Length > splitted[0].Length
&& !splitSequences.Contains(splitted[1]))
{
result.Add(splitted[0]);
result.AddRange(SplitInSegments(splitted[1], 2));
}
else
{
result.AddRange(SplitInSegments(splitted[0], 2));
result.Add(splitted[1]);
}
}
else // splitted.Count >= 3
{
// 3 segments is good.
result = splitted;
// More than 3 segments, combine some together.
while (result.Count > 3)
{
// Find the shortest combination of two segments
int shortestComboCount = int.MaxValue;
int shortestComboIndex = 0;
for (int i = 0; i < result.Count - 1; i++)
{
int currentComboCount = result[i].Length + result[i + 1].Length;
if (currentComboCount < shortestComboCount)
{
shortestComboCount = currentComboCount;
shortestComboIndex = i;
}
}
// Combine the shortest segments and replace in the result.
string combo = result[shortestComboIndex] + result[shortestComboIndex + 1];
result.RemoveAt(shortestComboIndex + 1);
result[shortestComboIndex] = combo;
}
}
return result;
}
现在当你调用代码时:
// always returns three segments.
var splitted = SplitWord(word);