Question

给定两个字符串text和pattern，找到text中匹配pattern的最短子字符串的开始和结束索引，这意味着pattern中的所有字符1}}在子字符串和pattern中以相同的顺序出现，但这些字符之间可能还有其他字符。

如果你能从text找到这样的子串，则打印它的开始和结束索引，否则打印-1，-1。如果有多个最短匹配子字符串，则返回具有最小开始索引的子字符串索引。

示例输入：

axxxbcaxbcaxxbc abc

abcd x

axxxbaxbab ab

示例输出：

有没有人有一些好的算法来解决这个问题而不使用内置支持C ++或Python中的正则表达式

Answer 1

循环文本的字符，找到文本中图案的第一个字符。如果找到它，在剩余的文本中搜索模式的第二个字符的出现，并在模式中的所有字符上重复操作，跳过文本中不需要的字符。完成后，再次启动下一个模式的第一个char文本。

使用abc模式可能更具视觉效果：

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

或者

 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，因为它是您正在寻找的子字符串的位置。

Answer 2

的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;
}

找到与另一个模式匹配的字符串中最短子字符串的开始和结束索引

2 个答案:

的Python

C ++