题目描述

s1,s2在s中最大跨距

(暴力枚举) $O(n^2)(doge)$

直接上暴力美学，一个从左边搜，一个从右边搜，搜到就break，然后输出距离

C++ 代码

#include<bits/stdc++.h>
using namespace std;
int main(){
    int f = 0, l = -5, r = -5;
    string ss, s = "", s1 = "", s2 = "";
    cin >> ss;
    for(int i = 0; i < ss.size(); i ++){
        if(ss[i] == ','){
            f ++;
            continue;
        }
        if(f == 0)s += ss[i];
        else if(f == 1)s1 += ss[i];
        else s2 += ss[i];
    }
    for(int i = 0; i < s.size(); i ++){
        string s3 = s.substr(i, s1.size());
        if(s3 == s1){l = i + s1.size();break;}
    }
    for(int i = s.size() - 1; i >= 0; i --){
        string s4 = s.substr(i - s2.size() + 1, s2.size());
        if(s4 == s2){r = i - s2.size();break;}
    }
    if(l == -5 || r == -5 || r - l <= 0)cout << -1 <<endl;
    else cout << r - l + 1 << endl;
    return 0;
}