Introduction to Algorithms: Palindrome substring (manacher) algorithm, O (n) Time efficiency realization

Problem Description:

Enter a string to find the largest palindrome substring in the list. The meaning of a substring is: a string fragment that appears consecutively in the original string. The meaning of a palindrome is: look and look backwards, like Abba and Yyxyy.

Analytical:

The O (n) palindrome string (manacher) algorithm is introduced here

The basic point of the algorithm: first, in a very ingenious way, all possible odd/even length palindrome substrings are converted to odd lengths :

Inserts a special symbol on both sides of each character. For example, Abba becomes #a #b#b#a#, ABA becomes #a #b#a#.

To further reduce the complexity of the encoding, you can add another special character at the beginning of the string, so that you do not have to deal with cross-border issues specifically, such as $ #a #b#a#.
The following is an example of string 12212321, which, after the previous step, became s[] = "$ #1 #2#2#1#2#3#2#1#";

Then use an array of p[i] to record the length (including S[i]) of the longest palindrome string centered on the character S[i], such as the corresponding relationship of S and P:

S # 1 # 2 # 2 # 1 # 2 # 3 # 2 # 1 #
P 1 2 1 2 5 2 1 4 1 2 1 6 1 2 1 2 1
(as you can see,p[i]-1 is exactly the total length of the palindrome string in the original P.S)

The following calculation P[i], the algorithm adds two auxiliary variable ID and MX, its ID represents the largest palindrome substring center position, MX is id+p[id], that is, the boundary of the largest palindrome substring.

The key point of this algorithm is here: if mx > I, then p[i] >= MIN (p[2 * id-i], mx-i).

The specific code is as follows:

if (mx > i) {      p[i] = (P[2*id-i] < (mx-i) P[2*id-i]: (Mx-i));} else{       P[i] = 1;}

When Mx-i > P[j], the palindrome string centered on S[j] is contained in a palindrome string centered on S[id], because I and J are symmetric, the palindrome string centered on s[i] is necessarily contained in a palindrome string centered on S[id], so there must be p[i] = P[j], see.

When P[j] > mx-i, the palindrome string centered on s[j] is not completely contained in a palindrome string centered on S[id], but based on symmetry, it is known that the part surrounded by the two green boxes is the same, that is, the palindrome string centered on the s[i], and its right will expand to the position of MX at least. , which means p[i] >= mx-i. As to whether the post-MX part is symmetrical, it is only one match.

For the case of MX <= i, p[i] can not be made more assumptions, only p[i] = 1, and then to match the

The following gives the original text, further explaining the reason for the linearity of the algorithm

Source:

#include <iostream> #include <string> #include <cstring>using namespace std;void findbmstr (string    & str) {int *p = new Int[str.size () + 1];    memset (p, 0, sizeof (p));    int mx = 0, id = 0;  for (int i = 1; I <= str.size (); i++) {if (mx > i) {p[i] = (P[2*id-i] < (mx-i)?        P[2*id-i]: (mx-i));        } else {p[i] = 1;        } while (str[i-p[i]] = = Str[i + p[i]]) p[i]++;            if (i + p[i] > mx) {mx = i + p[i];        id = i;    }} int max = 0, ii;            for (int i = 1; i < str.size (); i++) {if (P[i] > max) {II = i;        max = P[i];    }} max--;    int start = Ii-max;    int end = II + max;        for (int i = start; I <= end; i++) {if (str[i]! = ' # ') {cout << str[i];    }} cout << Endl; Delete p;} int main () {string str = "12212321";    String str0;    STR0 + = "$#";        for (int i = 0; i < str.size (); i++) {str0 + = Str[i];    STR0 + = "#";    } cout << str0 << Endl;    Findbmstr (STR0); return 0;}

Execution Result:

Introduction to Algorithms: Palindrome substring (manacher) algorithm, O (n) Time efficiency realization