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Recursive function for substring search in C++

给定两个字符串 Str 和 subStr 作为输入。目标是确定 subStr 中存在的文本是否作为子字符串存在于 Str 中。如果整个 X 在 Y 中至少出现一次,则字符串 X 称为 Y 的子串。 我们将使用递归方法来执行此操作。

例如

输入− Str = “tutorialspoint” subStr=”Point”

输出− 给定字符串不包含子字符串!

解释− 字符串 Point 不是教程点的子字符串

输入− Str = “globalization” subStr=”global”

输出− 给定字符串包含子字符串!

解释 - 字符串global是全球化的子字符串

下面的程序中使用的方法如下

在这种方法中,我们以递归方式检查subStr是否是Str的子字符串。递归的步骤为:-

  • 1.将两个字符串传递给递归函数,其中指针将指向两个字符串的当前字符位置

  • 如果字符串结束但模式还剩下更多字符,则返回 0,因为未找到模式我们到达了字符串的末尾。

  • 如果当前字符是模式中的最后一个字符,则在字符串中找到它,返回 1。

    li>
  • 如果两个当前字符相同,则将两个指针移动到下一个位置。

  • 如果两个当前字符不匹配,则将指针移动到下一个位置。将字符串移动到下一个位置。

  • 将输入字符串作为字符数组 Str 和 subStr。

  • 函数 match(char *str1, char *substr1) 接受两个子字符串,如果 substr1 和 str1 相同,则返回 1。

  • 两个指针都指向字符串中存在的字符,最初位于起始位置。

  • 如果 substr 为空,则返回 0。

  • 如果两个字符串都为空,则也返回 0。

  • 如果两个字符串都为空,则也返回 0。 >

  • 如果两个当前字符相等,则使用 match(str1 + 1, substr1 + 1) 递归检查下一个字符

  • 函数 checksubString (char *str2, char *substr2) 接受两个字符串,如果 str2 中存在 substr2,则返回 1。

  • 如果 str2 和 substr2 指向的当前字符相同,则检查是否连续的字符也可以使用 match() 函数进行匹配。如果返回 1,则返回 1。

  • 如果到达 str2 的末尾,则返回 0。

  • 否则递归检查str2 的下一个字符使用 checksubString(str2 + 1, substr2);

  • 如果所有条件都失败,则还使用递归检查 checksubString(str2 + 1, substr2);

  • 根据返回值打印结果。

示例

#include<iostream>
using namespace std;
int match(char *str1, char *substr1){
   if (*substr1 == &#39;\0&#39;){
      return 1;
   }
   if (*str1 == &#39;\0&#39;){
      if(*substr1 != &#39;\0&#39;){
         return 0;
      }
   }
   if (*str1 == *substr1){
      return match(str1 + 1, substr1 + 1);
   }
   return 0;
}
int checksubString(char *str2, char *substr2){
   if (*str2 == *substr2){
      if(match(str2, substr2)){
         return 1;
      }
   }
   if (*str2 == &#39;\0&#39;){
      return 0;
   }
   else{
      return checksubString(str2 + 1, substr2);
   }

   return checksubString(str2 + 1, substr2);
}
int main(){
   char Str[]="tutorialspoint";
   char subStr[]="point";

   if(checksubString(Str,subStr)==1){
      cout << "Given string contains substring!";
   }
   else{
      cout << "Given string does not contain substring!"; }
   return 0;
}

输出

如果我们运行上面的代码,它将生成以下输出

Given string contains substring!

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