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 == '\0'){ return 1; } if (*str1 == '\0'){ if(*substr1 != '\0'){ 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 == '\0'){ 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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