题解

之前已经做过详细的笔记了，有点类似于DP的思路

C++ 代码

#include <iostream>
using namespace std;
const int N = 100010;
const int M = 1000010;

int ne[N];

char p[N],s[M];

int main(){
    int n , m;
    cin >> n >> p + 1 >> m >> s + 1; //char数组从下标1开始
    //求next数组
    for(int i = 2 , j = 0;i <= n;++i){
        while(j && p[j + 1] != p[i]) j = ne[j];
        if(p[j + 1] == p[i]) ++j;
        ne[i] = j;
    }

    //开始匹配
    for(int i = 1,j = 0;i <= m;++i){
        while(j && p[j + 1] != s[i]) j = ne[j];
        if(p[j + 1] == s[i]) j ++;
        if(j == n){
            cout << i - n << " ";
            j = ne[j];
        }
    }
    return 0;
}