Manacher算法,最长回文串

时间:2023-03-09 18:38:30
Manacher算法,最长回文串

给你10000长度字符串,然你求最长回文字串,输出长度,暴力算法肯定超时

#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
using namespace std;

void findBMstr(string& str){
    int p[str.size()+1];
    memset(p, 0, sizeof(p));
    int mx = 0, id = 0;
    for(int i = 1; i <=  str.size(); i++){
        if(mx > i){
            //i的对称点的右边界是否过mx界,没有就取其对称点的值,否则就取当前可取的最大值(mx-i)
            p[i] = (p[2*id - i] < (mx - i) ? p[2*id - i] : (mx - i));
        }else{
            //意思是当前的点为止,就赋值为1
            p[i] = 1;
        }
        while(str[i - p[i]] == str[i + p[i]]){
            p[i]++;//一直向两边拓展
        }
        if(i + p[i] > mx){//更新id值为边界大的那个
            mx = i + p[i];
            id = i;
        }
    }
    printf("\n");
    int max = 0, ii;
    for(int i = 1; i < str.size(); i++){
        if(p[i] > max){//寻找最值,以及索引
            ii = i;
            max = p[i];
        }
    }

    max--;
    cout << max << endl;
}

int main()
{
    string str;
    cin >> str;
    string str0;
    str0 += "$#";
    for(int i = 0; i < str.size(); i++){
        str0 += str[i];
        str0 += "#";
    }
    findBMstr(str0);
    return 0;
}