Question

我不知何故需要在其他字符串中找到最长的字符串，所以如果string1为“Alibaba”而string2为“ba”，则最长的字符串为“baba”。我有字符串的长度，但下一步是什么？

char* fun(char* a, char& b) 
{
int length1=0;
int length2=0;
int longer;
int shorter;
char end='\0';

while(a[i] != tmp)
{
    i++;
    length1++;
}

int i=0;
while(b[i] != tmp)
{
    i++;
    length++;
}

if(dlug1 > dlug2){
    longer = length1;
    shorter = length2;
}
else{
    longer = length2;
    shorter = length1;
}

 //logics here
   }

int main()
{
char name1[] = "Alibaba";
char name2[] = "ba";
char &oname = *name2;

cout << fun(name1, oname) << endl;

system("PAUSE");
return 0;
}

Answer 1

哇这个问题的答案很多。以下是您的代码应该执行的操作：

使用标准字符串搜索功能查找“ba”的第一个实例。
在循环中查看此“ba”以查看接下来N个字符中有多少也是“ba”。
如果此序列长于先前记录的最长序列，请保存其长度和位置。
在最后一个之后找到下一个“ba”的实例。

这是代码（未经测试）：

string FindLongestRepeatedSubstring(string longString, string shortString)
{
    // The number of repetitions in our longest string.
    int maxRepetitions = 0;

    int n = shortString.length(); // For brevity.

    // Where we are currently looking.
    int pos = 0;
    while ((pos = longString.find(shortString, pos)) != string::npos)
    {
        // Ok we found the start of a repeated substring. See how many repetitions there are.
        int repetitions = 1;
        // This is a little bit complicated.
        // First go past the "ba" we have already found (pos += n)
        // Then see if there is still enough space in the string for there to be another "ba"
        // Finally see if it *is* "ba"
        for (pos += n; pos+n < longString.length() && longString.substr(pos, n) == shortString; pos += n)
            ++repetitions;
        // See if this sequence is longer than our previous best.
        if (repetitions > maxRepetitions)
            maxRepetitions = repetitions;
    }
    // Construct the string to return. You really probably want to return its position, or maybe
    // just maxRepetitions.
    string ret;
    while (maxRepetitions--)
        ret += shortString;
    return ret;
}

Answer 2

http://www.cplusplus.com/reference/string/string/find/

也许你是故意这样做的，但是你应该使用std :: string类并忘记像char * string表示这样的古老的东西。它将使您能够使用许多优化方法，如字符串研究等。

Answer 3

你想要的应该看起来像这个伪代码：

i = j = count = max = 0
while (i < length1 && c = name1[i++]) do
  if (j < length2 && name2[j] == c) then
    j++
  else
    max = (count > max) ? count : max
    count = 0
    j = 0
  end
  if (j == length2) then
    count++
    j = 0
  end
done
max = (count > max) ? count : max
for (i = 0 to max-1 do
  print name2
done

这个想法在这里，但我觉得可能存在一些这种算法不起作用的情况（具有复杂重叠的情况需要返回name1）。你可能想看一下Boyer-Moore算法并将两者混合起来得到你想要的东西。

Answer 4

Algorithms Implementation Wikibook可以在C ++中实现您想要的功能。

Answer 5

为什么不使用C提供的strstr函数。

const char * strstr ( const char * str1, const char * str2 );
      char * strstr (       char * str1, const char * str2 );
Locate substring

Returns a pointer to the first occurrence of str2 in str1,
or a null pointer if str2 is not part of str1.
The matching process does not include the terminating null-characters.

现在使用长度并创建一个循环并使用原始字符串播放anf找到里面最长的字符串。

如何在字符串中查找字符串

5 个答案: