Question

说我有两个字符串：

"SomeTextHereThatIsTheSource"
"SomeTextHereThatIsCloseToTheSourceButNotTheSame"

是否有一种灵巧的.net方式来了解文本的哪一部分是相同的（从头开始）。

所以在这个例子中结果是：

"SomeTextHereThatIs"

Answer 1

你可以使用TakeWhile：

string MatchFromStart(string s1, string s2)
{
    if (s1 == null || s2 == null) return "";

    var matchingArray = s1.TakeWhile((c, i) => { return i < s2.Length && c == s2[i]; });
    return String.Join("", matchingArray);
}

然后使用它：

string s1 = "SomeTextHereThatIsTheSource";
string s2 = "SomeTextHereThat";
string s3 = "SomeTextHereThatIsCloseToTheSourceButNotTheSame";
Console.WriteLine(MatchFromStart(s1, s2));   // SomeTextHereThat
Console.WriteLine(MatchFromStart(s2, s1));   // SomeTextHereThat
Console.WriteLine(MatchFromStart(s3, s1));   // SomeTextHereThatIs
Console.WriteLine(MatchFromStart("", s1));   // (blank string)
Console.WriteLine(MatchFromStart(s3, ""));   // (blank string)
Console.WriteLine(MatchFromStart(null, s1)); // (blank string)
Console.WriteLine(MatchFromStart(s2, null)); // (blank string)

Answer 2

我会在较短的字符串上说for循环，逐个字符地比较是最好的选择。它也可能是最快的。

// str1 is shorter or equal in length to str2:
for(int i=0; i < str1.Length; i++)
{
  if(str1[i] == str2[i])
   continue;

  return i;
}

Answer 3

使用简单的while循环来确定匹配的长度：

int len = 0;
while (len < s1.Length && len < s2.Length && s1[len] == s2[len]) {
    len++;
}

如果您事先确定两个字符串的最小长度，可以稍微改进一下

int minLength = Math.Min(s1.Length, s2.Length);
int len = 0;
while (len < minLength && s1[len] == s2[len]) {
    len++;
}

Answer 4

虽然它可能不是最优化的，但这是一个非常干净的方法来利用string也是IEnumerable<char>的事实。

static string CommonPrefix(string string1, string string2) {
    var length = Enumerable.Zip(string1, string2, (a, b) => a == b).TakeWhile(a => a).Count();
    return string1.Substring(0, length);
}

找出字符串匹配的数量

4 个答案: