(algorithm) the longest palindrome substring

Topic:

To find the longest palindrome substring of a string

Ideas:

1. Violent enumeration

The most easy to think of is brute force, enumerate each substring, and then according to the definition of palindrome is not a palindrome, find the longest one.

The time complexity of each substring is O (n^2) and the time complexity of the substring is O (N), so the time complexity is O (n^3).

2. Dynamic planning

The substring of the back character string is also a palindrome, such as p[i,j] (indicating that the substring ending with J begins with I) is a palindrome string, then P[i+1,j-1] is also a palindrome string. So the longest palindrome string can be decomposed into a series of sub-problems.

This requires an extra space of O (n^2), and the time complexity is O (n^2).

State transition equation:

When Str[i]=str[j] dp[i][j]=dp[i+1][j-1]

When Str[i]!=str[j] Dp[i][j]=0

Initial state:

Dp[i][i]=1

Dp[i][i+1]=1 if STR[I]=STR[I+1]

3. Center expansion

The central extension is to take each character of a given string as a center and extend it to both sides to find the longest child palindrome. The time complexity is O (n^2).

But there are two things to consider:

such as ABA, this is an odd length.

such as ABBA, so that the length is even.

4. Manacher method

Cond

Code:

#include <iostream>using namespace std;//Brute forcestring findlongestpalindrome_1 (string &str) {int length= Str.size (); int maxlength=0;int start;for (int i=0;i<length;i++) {for (int j=i+1;j<length;j++) {int left,right;for (left=i,right=j;left<right;left++,right--) {if (str[left]!=str[right]) break;} if (Left>=right && (j-i) >maxlength) {maxlength=j-i+1;start=i;}}} if (maxlength>0) return str.substr (start,maxlength); return NULL;} Dynamic programmingstring findlongestpalindrome_2 (string &str) {const int length=str.size (); int Maxlength=1;int Start;bool dp[length][length];for (int i=0;i<length;i++) for (int j=0;j<length;j++) dp[i][j]=false;for (int i=0;i <length;i++) {dp[i][i]=true;if (i<length-1 && str[i]==str[i+1]) {dp[i][i+1]=true;start=i;maxlength=2;}} for (int len=3;len<=length;len++) {for (int i=0;i<=length-len;i++) {int j=i+len-1;if (dp[i+1][j-1] && str[ I]==str[j]) {Dp[i][j]=true;start=i;maxlength=len;}}} if (maxlength>=2) return str.substr(start,maxlength); return NULL;} Pivot expandstring findlongestpalindrome_3 (string &str) {const int length=str.size (); int maxlength=1;int start; for (int i=0;i<length;i++) {int Left=i-1;int right=i+1;while (left>=0 && right<=length-1 && str [Left]==str[right]) {if (right-left+1>maxlength) {start=left;maxlength=right-left+1;} left--;right++;}} for (int i=0;i<length;i++) {int Left=i;int right=i+1;while (left>=0 && right<=length-1 && str[ Left]==str[right]) {if (right-left+1>maxlength) {start=left;maxlength=right-left+1;} left--;right++;}} if (maxlength>0) return str.substr (start,maxlength); return NULL;} int main () {string str= "ABABCDDCBBD"; cout<<findlongestpalindrome_1 (str) <<endl;cout<< Findlongestpalindrome_2 (str) <<endl;cout<<findlongestpalindrome_3 (str) <<endl;return 0;}

　　

(algorithm) the longest palindrome substring