我正在使用C＃来查找两个字符串之间的所有常见子字符串。例如，如果输入是： S1 =“需要电子邮件协商” S2 =“需要电子邮件协助”
输出应该是“需要帮助电子邮件”
下面的代码返回最长的公共子字符串，但我希望我的代码返回所有常见的子字符串。任何帮助深表感谢！
static void commonsubstrings()
{
input1 = "need asssitance with email";
input2 = "email assistance needed"
if (input2.Length > input1.Length)
{
swap = input1;
input1 = input2;
input2 = swap;
}
int k = 1;
String temp;
String longTemp = "";
for (int i = 0; (i <= input1.Length); i++)
{
if ((i == input1.Length))
{
if (longest != null)
{
k = longest.Length + 1;
}
else
{
k = 1;
}
temp = input1.Substring(1, input1.Length - 1);
if (temp.Equals(""))
{
break;
}
if (k <= temp.Length)
{
i = k - 1;
input1 = temp;
if ((longest != null) && (longest.Length > longTemp.Length))
{
longTemp = longest;
}
}
}
holder1 = input1.Substring(0, k);
for (int j = 0; (j < input2.Length) && (j + k <= input2.Length); j++)
{
check1 = input2.Substring(j, k);
if (holder1.Equals(check1))
{
longest = holder1;
break;
}
}
k++;
}
Console.WriteLine(longest);
Console.ReadLine();
}
答案 0 :(得分：4)
另一种方法：您可以使用levenshtein distance来查找两个单词的相似度。如果距离小于指定值，您可以认为两个字符串相等。然后你可以使用levenshtein比较器Enumerable.Intersect。
然后很简单：
string S1= "need asssitance with email" ;
string S2 = "email assistance needed";
string[] words1 = S1.Split();
string[] words2 = S2.Split();
var wordsIntersecting = words1.Intersect(words2, new LevenshteinComparer());
string output = string.Join(" ", wordsIntersecting);
输出：need asssitance email
这是自定义比较器：
class LevenshteinComparer : IEqualityComparer<string>
{
public int MaxDistance { get; set; }
private Levenshtein _Levenshtein = new Levenshtein();
public LevenshteinComparer() : this(50) { }
public LevenshteinComparer(int maxDistance)
{
this.MaxDistance = maxDistance;
}
public bool Equals(string x, string y)
{
int distance = _Levenshtein.iLD(x, y);
return distance <= MaxDistance;
}
public int GetHashCode(string obj)
{
return 0;
}
}
这是levenshtein算法的实现：
public class Levenshtein
{
///*****************************
/// Compute Levenshtein distance
/// Memory efficient version
///*****************************
public int iLD(String sRow, String sCol)
{
int RowLen = sRow.Length; // length of sRow
int ColLen = sCol.Length; // length of sCol
int RowIdx; // iterates through sRow
int ColIdx; // iterates through sCol
char Row_i; // ith character of sRow
char Col_j; // jth character of sCol
int cost; // cost
/// Test string length
if (Math.Max(sRow.Length, sCol.Length) > Math.Pow(2, 31))
throw (new Exception("\nMaximum string length in Levenshtein.iLD is " + Math.Pow(2, 31) + ".\nYours is " + Math.Max(sRow.Length, sCol.Length) + "."));
// Step 1
if (RowLen == 0)
{
return ColLen;
}
if (ColLen == 0)
{
return RowLen;
}
/// Create the two vectors
int[] v0 = new int[RowLen + 1];
int[] v1 = new int[RowLen + 1];
int[] vTmp;
/// Step 2
/// Initialize the first vector
for (RowIdx = 1; RowIdx <= RowLen; RowIdx++)
{
v0[RowIdx] = RowIdx;
}
// Step 3
/// Fore each column
for (ColIdx = 1; ColIdx <= ColLen; ColIdx++)
{
/// Set the 0'th element to the column number
v1[0] = ColIdx;
Col_j = sCol[ColIdx - 1];
// Step 4
/// Fore each row
for (RowIdx = 1; RowIdx <= RowLen; RowIdx++)
{
Row_i = sRow[RowIdx - 1];
// Step 5
if (Row_i == Col_j)
{
cost = 0;
}
else
{
cost = 1;
}
// Step 6
/// Find minimum
int m_min = v0[RowIdx] + 1;
int b = v1[RowIdx - 1] + 1;
int c = v0[RowIdx - 1] + cost;
if (b < m_min)
{
m_min = b;
}
if (c < m_min)
{
m_min = c;
}
v1[RowIdx] = m_min;
}
/// Swap the vectors
vTmp = v0;
v0 = v1;
v1 = vTmp;
}
// Step 7
/// Value between 0 - 100
/// 0==perfect match 100==totaly different
///
/// The vectors where swaped one last time at the end of the last loop,
/// that is why the result is now in v0 rather than in v1
//System.Console.WriteLine("iDist=" + v0[RowLen]);
int max = System.Math.Max(RowLen, ColLen);
return ((100 * v0[RowLen]) / max);
}
///*****************************
/// Compute the min
///*****************************
private int Minimum(int a, int b, int c)
{
int mi = a;
if (b < mi)
{
mi = b;
}
if (c < mi)
{
mi = c;
}
return mi;
}
///*****************************
/// Compute Levenshtein distance
///*****************************
public int LD(String sNew, String sOld)
{
int[,] matrix; // matrix
int sNewLen = sNew.Length; // length of sNew
int sOldLen = sOld.Length; // length of sOld
int sNewIdx; // iterates through sNew
int sOldIdx; // iterates through sOld
char sNew_i; // ith character of sNew
char sOld_j; // jth character of sOld
int cost; // cost
/// Test string length
if (Math.Max(sNew.Length, sOld.Length) > Math.Pow(2, 31))
throw (new Exception("\nMaximum string length in Levenshtein.LD is " + Math.Pow(2, 31) + ".\nYours is " + Math.Max(sNew.Length, sOld.Length) + "."));
// Step 1
if (sNewLen == 0)
{
return sOldLen;
}
if (sOldLen == 0)
{
return sNewLen;
}
matrix = new int[sNewLen + 1, sOldLen + 1];
// Step 2
for (sNewIdx = 0; sNewIdx <= sNewLen; sNewIdx++)
{
matrix[sNewIdx, 0] = sNewIdx;
}
for (sOldIdx = 0; sOldIdx <= sOldLen; sOldIdx++)
{
matrix[0, sOldIdx] = sOldIdx;
}
// Step 3
for (sNewIdx = 1; sNewIdx <= sNewLen; sNewIdx++)
{
sNew_i = sNew[sNewIdx - 1];
// Step 4
for (sOldIdx = 1; sOldIdx <= sOldLen; sOldIdx++)
{
sOld_j = sOld[sOldIdx - 1];
// Step 5
if (sNew_i == sOld_j)
{
cost = 0;
}
else
{
cost = 1;
}
// Step 6
matrix[sNewIdx, sOldIdx] = Minimum(matrix[sNewIdx - 1, sOldIdx] + 1, matrix[sNewIdx, sOldIdx - 1] + 1, matrix[sNewIdx - 1, sOldIdx - 1] + cost);
}
}
// Step 7
/// Value between 0 - 100
/// 0==perfect match 100==totaly different
System.Console.WriteLine("Dist=" + matrix[sNewLen, sOldLen]);
int max = System.Math.Max(sNewLen, sOldLen);
return (100 * matrix[sNewLen, sOldLen]) / max;
}
}
对Levenshtein课程的信任：http://www.codeproject.com/Articles/13525/Fast-memory-efficient-Levenshtein-algorithm
答案 1 :(得分：3)
public static string [] CommonString(string left, string right)
{
List<string> result = new List<string>();
string [] rightArray = right.Split();
string [] leftArray = left.Split();
result.AddRange(rightArray.Where(r => leftArray.Any(l => l.StartsWith(r))));
// must check other way in case left array contains smaller words than right array
result.AddRange(leftArray.Where(l => rightArray.Any(r => r.StartsWith(l))));
return result.Distinct().ToArray();
}
答案 2 :(得分：1)
使用Set-Intersections
从例程开始，查找字符串的所有可能的子字符串。这是在Python中，它是“读者练习”将其翻译为C＃'：
def allSubstr(instring):
retset = set()
retset.add(instring)
totlen = len(instring)
for thislen in range(0, totlen):
for startpos in range(0, totlen):
# print "startpos: %s, thislen: %s" % (startpos, thislen)
addStr = instring[startpos:startpos+thislen]
# print "addstr: %s" % (addStr)
retset.add(addStr)
print "retset total: %s" % (retset)
return retset
set1 = allSubstr('abcdefg')
set2 = allSubstr('cdef')
print set1.intersection(set2)
这是输出：
>>> set1 = allSubstr('abcdefg')
retset total: set(['', 'cde', 'ab', 'ef', 'cd', 'abcdef', 'abc', 'efg', 'bcde', 'cdefg', 'bc', 'de', 'bcdef', 'abcd', 'defg', 'fg', 'cdef', 'a', 'c', 'b', 'e', 'd', 'g', 'f', 'bcd', 'abcde', 'abcdefg', 'bcdefg', 'def'])
>>> set2 = allSubstr('cdef')
retset total: set(['', 'cde', 'c', 'ef', 'e', 'd', 'f', 'de', 'cd', 'cdef', 'def'])
>>>
>>> set1.intersection(set2)
set(['', 'cde', 'c', 'de', 'e', 'd', 'f', 'ef', 'cd', 'cdef', 'def'])
不，你对长度为1的子集不感兴趣。但是，在进行set.add（）调用之前，你总是可以添加一个长度限制。