KMP string pattern matching

The function used here is from the leetcode. Details can be found in leetcode problem: Implement strStr()

 

The best explanation should be made in the comments, which can be understood by the leading of code.

// next[j]: the smallest valid position we need to check next when detect mismatch at jth character pattern[j]
// Here, valid position means "pattern[0, ..., next[j]-1]" are matched with "text".

void getNext(char *pattern, int next[]){
	int i = 0, j = -1;
	
	// If the "text[i]" fails to match "patter[0]", then we need to 
	// check "text[i+1]" and "patter[0]", which also means "text[i]"
	// would check with "patter[-1]".
	next[i] = j;	 
	
	// while loop 1:
	while(pattern[i] != '\0){
			// while loop 2:
			while(j >= 0 && pattern[i] != pattern[j]){
				// First, j need to be valid index, so it needs to be not less than 0.
				// Then, if "pattern[i]" fails to match "pattern[j]", we can also think as
				// "text[i]" fails to match "pattern[j]".
				// So we need to check if "text[i]" matches with "pattern[next[j]]", as next[j]
				// is the position we need to check when we fail at position j.
				j = next[j];				
			}
			
			// After the above while loop, we can know that "text[0, ..., i]" matches 
			// "pattern[0, ..., j]", so we can move one more step for both "text" and "pattern".
			++i; ++j;
			
			// For the new i, marked as i_new, we can determine its "next value" now!!
			// As we've known that "text[0, ..., i_new - 1]" matches "pattern[0, ..., j_new - 1]",
			// if we fail to match at position "text[i_new]", we can move pattern to the j_new position to
			// check if "text[i_new]" matches "pattern[j_new]".
			// P.S:
			// Also, we can know the j_new position is the optimized position. If we can get a valid position j' (valid 
			// means "pattern[0, ..., j'-1]" are matched with "text") smaller
			// than j_new, then we'd get "(j' - 1)" (which is valid at position j'-1) is smaller than "next[j]", 
			// which is contradicted to the definition of "next" table.
			if(pattern[i] == pattern[j])
				next[i] = next[j];
			else
				next[i] = j;			
	}	
}


char *strStr(char *text, char *pattern){
	if(NULL == text || NULL == pattern)
		return NULL;
	if('\0' == pattern[0])
		return text;
	
	// i is the pointer of text, j is the pointer of pattern.
	int i = 0, j = 0;
	char *pos = NULL;
	int *next = new int[strlen(pattern) + 1];	// include the '\0'
	
	getNext(pattern, next);
	
	while(text[i] != '\0'){
		// Same optimization in getNext(), that is
		// if we fail at one position, we may also fail at the 
		// next position, which means we can continue along the "next" table
		// Also, we need the index to be valid first.
		while(j >= 0 && text[i] != pattern[j])
			j = next[j];
		
		// After the while loop, we can know "text[0, ..., i]" matches "pattern[0, ..., j]"
		// So we need to move one more step for both "text" and "pattern".
		++i; ++j;
		
		if(pattern[j] == '\0'){
			pos = (text + i) - j;	// The beginning position in text which corresponding to the matched pattern position.
			return pos;
		}		
	}
	
	return pos;
	
}

 

posted @ 2015-03-26 23:06  kid551  阅读(224)  评论(0编辑  收藏  举报