Z-box / Z-algorithm string matching

Brief introduction

z [i]: for i and the position of the original prefix string of the same length is the number

p [] + "#" + s [] - matching strings found for each text string of the same position and its prefix

Processing array z

Z [0] = 0 is not valid, with z [k] (k <i) to Release z [i]

Z look quite good before there can cover the current i, and the right boundary of the right cover better l, r is the boundary box

if (no previous zbox to cover) the enumeration process had to answer this one from i violence

else if (has) s [i] can be overwritten so s [i] is inevitable in the s [0 ~ z [j] - 1] j at the beginning of this section intended to cover the maximum current is maintained to the right boundary i

So we will now go to find the position s [i] is mapped in the original string, i.e., i - l

At this time Obviously, it s in the box range [i ~ r] and the mapping s [il ~ rl] is same length of this section is L = r - i + 1

It is easy to see two types of situation:

1.z [i - l] <L can be derived directly z [i] = z [i - l]

2. The excess part of the border had to find an enumeration of violence

 

template

// LEN-- text strings match string len--

void getz(){
	int l = 0,r = 0;
	z[0] = 0;
	for(int i = 1;i <= LEN + len; i++){
		if(i > r){
			int n = 0;
			while(p[n] == p[i + n]) n++;
			if(n){
				l = i;
				r = i + n - 1;
			}
			z[i] = n;
		}
		else{
			if(z[i - l] < r - i + 1) z[i] = z[i - l];
			else{
				int n = 1;
				while(p[r - i + n] == p[r + n]) n++;
				r = r + n - 1;
				l = i;
				z[i] = r - l + 1;
			}
		}
	}
}