在字符串C++中查找子字符串(在"hello"中查找"el")

Finding substring within string C++ (find "el" in "hello")

本文关键字：查找 字符串 el hello C++      更新时间：2023-10-16

好的，所以我正在寻找一种算法，可以帮助我在子字符串中找到字符串。我以前使用的代码来自一个类似的问题，但它不起作用

// might not be exposed publicly, but could be
int index_of(string const& haystack, int haystack_pos, string const& needle) {
// would normally use string const& for all the string parameters in this
// answer, but I've mostly stuck to the prototype you already have
// shorter local name, keep parameter name the same for interface clarity
int& h = haystack_pos;
// preconditions:
assert(0 <= h && h <= haystack.length());
if (needle.empty()) return h;
if (h == haystack.length()) return -1;
if (haystack.compare(h, needle.length(), needle) == 0) {
return h;
}
return index_of(haystack, h+1, needle);
}
int index_of(string haystack, string needle) {
// sets up initial values or the "context" for the common case
return index_of(haystack, 0, needle);
}

这不会返回字符串"hello"上的起始索引"el"，我也搞不清楚。

编辑：好吧，让我向您展示更多的代码，包括一些现实生活中的示例：我试图分析一个字符串，它是我想在文件系统中排序的文件的路径。输入示例如下：

输入：/media/seagate/lol/Sons.of.Archy.S04.720p.HDTV.x264/Sons.of-Archy.S04E01.720p.HDTV.xs264-IMMERSE.mkv

当我试图通过检测SxxExx的存在来解析这个字符串以获得的名称时，我会寻找"s0"、"s0"等(我知道这不是最好的实现，我只是想看看它是否有效，然后再看代码)。所以当我使用这个输入时，我在输出上得到的是：

input:/media/seagate/lol/Sons.of.Anarchy.S04.720p.HDTV.x264/Sons.of.Anarchy.S04E01.720p.HDTV.x264-IMMERSE.mkv
aux: 0p.HDTV.x264-IMMERSE.mkv
input:/media/seagate/lol/Sons.of.Anarchy.S04.720p.HDTV.x264/Sons.of.Anarchy.S04E01.720p.HDTV.x264-IMMERSE.mkv
aux: 1.720p.HDTV.x264-IMMERSE.mkv
input:/media/seagate/lol/Sons.of.Anarchy.S04.720p.HDTV.x264/Sons.of.Anarchy.S04E01.720p.HDTV.x264-IMMERSE.mkv
aux: 264-IMMERSE.mkv

aux的预期输出：S04E01.720p.HDTV.x264-IMMERSE.mkv

因此，正如你所看到的，它只是在寻找字符串中的任何字符并停止，这也解释了多个有效的"已找到"，而这些字符本应只是一个。

我尝试使用的完整代码是：

bool StringWorker::isSeries(size_t &i) {
size_t found1, found2, found3, found4, found5, found6;
found1 = input->find_last_of("S0"); //tried several find functions including the
found2 = input->find_last_of("S1"); //index_of() mentioned above in the post
found3 = input->find_last_of("S2");
found4 = input->find_last_of("s0");
found5 = input->find_last_of("s1");
found6 = input->find_last_of("s2");
if (found1 != string::npos) {
if (input->size() - found1 > 6) {
string aux = input->substr(found1, input->size());
cout << "input:" << *input << endl;
cout << "aux: " << aux << endl;
if (isalpha(aux.at(0)) && isdigit(aux.at(1)) && isdigit(aux.at(2))
&& isalpha(aux.at(3)) && isdigit(aux.at(4))
&& isdigit(aux.at(5))) {
i = found1;
return true;
}
}
}
if (found2 != string::npos) {
if (input->size() - found2 > 6) {
string aux = input->substr(found2, input->size());
cout << "input:" << *input << endl;
cout << "aux: " << aux << endl;
if (isalpha(aux.at(0)) && isdigit(aux.at(1)) && isdigit(aux.at(2))
&& isalpha(aux.at(3)) && isdigit(aux.at(4))
&& isdigit(aux.at(5))) {
i = found2;
return true;
}
}
}
if (found3 != string::npos) {
if (input->size() - found3 > 6) {
string aux = input->substr(found3, input->size());
cout << "input:" << *input << endl;
cout << "aux: " << aux << endl;
if (isalpha(aux.at(0)) && isdigit(aux.at(1)) && isdigit(aux.at(2))
&& isalpha(aux.at(3)) && isdigit(aux.at(4))
&& isdigit(aux.at(5))) {
i = found3;
return true;
}
}
}
if (found4 != string::npos) {
if (input->size() - found4 > 6) {
string aux = input->substr(found4, input->size());
cout << "input:" << *input << endl;
cout << "aux: " << aux << endl;
if (isalpha(aux.at(0)) && isdigit(aux.at(1)) && isdigit(aux.at(2))
&& isalpha(aux.at(3)) && isdigit(aux.at(4))
&& isdigit(aux.at(5))) {
i = found4;
return true;
}
}
}
if (found5 != string::npos) {
if (input->size() - found5 > 6) {
string aux = input->substr(found5, input->size());
cout << "input:" << *input << endl;
cout << "aux: " << aux << endl;
if (isalpha(aux.at(0)) && isdigit(aux.at(1)) && isdigit(aux.at(2))
&& isalpha(aux.at(3)) && isdigit(aux.at(4))
&& isdigit(aux.at(5))) {
i = found5;
return true;
}
}
}
if (found6 != string::npos) {
if (input->size() - found6 > 6) {
string aux = input->substr(found6, input->size());
cout << "input:" << *input << endl;
cout << "aux: " << aux << endl;
if (isalpha(aux.at(0)) && isdigit(aux.at(1)) && isdigit(aux.at(2))
&& isalpha(aux.at(3)) && isdigit(aux.at(4))
&& isdigit(aux.at(5))) {
i = found6;
return true;
}
}
}
return false;
}

你看到这里有什么不对吗？

为什么不使用std::string->链接的find()方法呢。

此代码通过index = sub_str.find("el"):返回索引

#include <iostream>
#include <string>
using namespace std;
int main ()
{
string sub_str="abc def ghi jk lmnop  hello";
string sub_str2;
size_t index;

index = sub_str.find("el");     
sub_str2 = sub_str.substr (index);   
cout<<"index = "<<index<<"n";
cout<<sub_str2<<"n";
return 0;
}

要在字符串中查找子字符串及其索引，可以尝试以下方法-

int find_sub(const std::string& mstring,sub)
{
int lensub=sub.length(),len=mstring.length(),f=0,pos;
std::string b="";
for(int i=0;i<len-lensub;i++)
{
for(int j=i,k=0;j<i+lensub;j++,k++)
b[k]=mstring[j];
if(b.compare(sub)==0)
{
f=1;
pos=i;
break;
}
} 
if(f==1)
cout<<"substring found at: "<<pos+1;
else
cout<<"substring not found!";
return f;
}

您还可以通过删除break;并每次增加f的值来检查子字符串出现的次数。还可以通过将pos转换为数组来获取它们的索引。

相关文章：