在匹配另一个模式的字符串中找到最短子字符串的开始和结尾索引

python

def shortest_match(text, pattern):
    stack = [] # to store matches
    for i in range(len(text) - len(pattern) + 1):
        # if we match the firts character of pattern in
        # text then we start to search for the rest of it
        if pattern[0] == text[i]:
            j = 1 # pattern[0] already match, let's check from 1 onwards
            k = i + 1 # text[i] == pattern[0], let's check from text[i+1] onwards
            # while pattern[j] could match text[i]
            while j < len(pattern) and k < len(text):
                if pattern[j] == text[k]:
                    j += 1 # pattern[j] matched. Let's move to the next character
                k += 1
            if j == len(pattern): # if the match was found add it to the stack
                stack.append((i, k-1))
            else: # otherwise break the loop (we won't find any other match)
                break
    if not stack: # no match found
        return (-1, -1)
    lengths = [y - x for x, y in stack] # list of matches lengths
    return stack[lengths.index(min(lengths))] # return the shortest

C

#include <iostream>
#include <vector>
#include <string.h>
using namespace std;
struct match_pair
{
    int start;
    int end;
    int length;
};
void
print_match (match_pair m)
{
    cout << "(" << m.start << ", " << m.end << ")";
}
match_pair 
shortest_match (char * text, char * pattern) 
{
  vector <match_pair> stack; // to store matches
  for (int i = 0; strlen(text) - strlen(pattern) + 1; ++i)
  {
    // if we match the firts character of pattern in
    // text then we start to search for the rest of it
    if (pattern[0] == text[i])
    {
        int j = 1; // pattern[0] already match, let's check from 1 onwards
        int k = i + 1; // text[i] == pattern[0], let's check from text[i+1] onwards
        // while pattern[j] could match text[i]
        while (j < strlen(pattern) && k < strlen(text))
        {
            if (pattern[j] == text[k])
            {
                ++j; // pattern[j] matched. Let's move to the next character
            }
            ++k;
        }
        if (j == strlen(pattern)) // if the match was found add it to the stack
        {
            match_pair current_match;
            current_match.start = i;
            current_match.end = k - 1;
            current_match.length = current_match.end - current_match.start;
            stack.push_back(current_match);
        } else // otherwise break the loop (we won't find any other match)
        {
            break;
        }
    }
  }
  match_pair shortest;
  if (stack.empty()) // no match, return (-1, -1)
  {
    shortest.start = -1;
    shortest.end = -1;
    shortest.length = 0;
    return shortest;
  }
  // search for shortest match
  shortest.start = stack[0].start;
    shortest.end = stack[0].end;
    shortest.length = stack[0].length;
  for (int i = 1; i < stack.size(); ++i)
  {
    if (stack[i].length < shortest.length)
    {
        shortest.start = stack[i].start;
        shortest.end = stack[i].end;
        shortest.length = stack[i].length;
    }
  }
  return shortest;
}
// override << for printing match_pair
std::ostream& 
operator<< (std::ostream& os, const match_pair& m)
{
    return os << "(" <<  m.start << ", " << m.end << ")"; 
}
int
main () 
{
  char text[] = "axxxbcaxbcaxxbc";
  char pattern[] = "abc";
  cout << shortest_match(text, pattern);
  return 0;
}

在文本的字符上循环，并在文本中找到图案的第一个字符。如果找到它，请在其余文本中搜索图案的第二个字符的情况，然后重复所有模式中的所有字符，然后在文本中跳过不必要的字符。完成后，重新开始，从文本中的模式的第一个字符开始。

也许使用abc模式更为视觉：

axxxbcaxbcaxxbc
[axxx|b|c] -> 6 chars
[ax|b|c] -> 4 chars
[axx|b|c] -> 5 chars

或

 aababaccccccc
[aa|baba|c] -> 6 chars
[a|baba|c] -> 5 chars
[a|ba|c] -> 4 chars
[accccccc] -> -1 chars as the substring does not match the pattern

编辑：您应该尝试从文本末尾开始实现此算法，因为这是您要寻找的子字符串的位置。