String application -- string violent matching (implemented in C language)

String application — String matching (C Language implementation )

Algorithmic thought – for instance

​ Just look at the picture ：

​ 【 Match success 】

​ ① At the beginning ：

​ ② The first one can match , Pointer backward ：

​ ③ The second one can also match , And the substring has reached the end , Match over , The match is successful .

​ 【 Matching failure 】

​ ① The first step is the same as above ： Both pointers point to the first character .

​ ② The second step , Pointer backward ：

​ In the second step, a mismatch is found , At this time , Need to update pointer i The location of , Because the first character has been compared , therefore i Should point to the 2 individual , The pointer j Go back to the beginning every time ：

​ ③ Because it doesn't match yet , therefore i Move all the way back , Until the end of the main string :

​ It has reached the end of the main string , Match over , But because of the mismatch , So the match failed .

​ Observe the law and you will find ：

​ ① If the characters match , be i++,j++;

​ ② If the characters do not match , be i = i - j +1; j = 0 .

​ When the match ends , if j The value of is equal to its length , Then the match is successful , Otherwise the match fails .

Code implementation

#include<stdio.h>
#include<stdbool.h> // according to C99 standard ,C Language use bool Type needs to add this header file 
#include <string.h>//strlen Function header file 

bool ForcedMatch(char* MainStr, char* SubStr)
{
    
	int i = 0;// Point to the main string 
	int j = 0;// Point to substring 
	
	while(i<strlen(MainStr)&&j<strlen(SubStr)){
    
		if(MainStr[i]==SubStr[j]){
    // If the current character matches  
			i++;
			j++;
		}
		else{
    	// If it doesn't match 
			i = i - j + 1;// Main string adjustment position  
			j = 0;// Violent matching , Substrings always go back to the beginning  
		}
	}
	// It has been matched to the end 
	return j==strlen(SubStr)?true:false;
} 

int main()
{
    
	printf(ForcedMatch("hello","he")?" The match is successful !\n":" Matching failure !\n");
	printf(ForcedMatch("hello","heo")?" The match is successful !\n":" Matching failure !\n");
	return 0;
}

test