小米算法第1道字符串，我也来提供一个思路

根据两个字符串的长度建立一个dp数组，相等的字符置1，然后遍历，时间复杂度有点高

#include<iostream>
#include<string>
#include<vector>
using namespace std;
void cal(string text, string pa, int & start, int & end){
    int len1=text.length();
    int len2=pa.length();
    vector<vector<int>> dp(len2,vector<int>(len1));
    for(int i=0;i<len2;i++){
        for(int j=0;j<len1;j++){
            if(text[j]!=pa[i]) dp[i][j]=0;
            else  dp[i][j]=1;
        }

    }
    start=-1;    end=-1;
    int sum=0,sumt=10000;

    for(int j=0;j<len1;j++){
        int i=0;
        int t=0;
        if(dp[i][j]==1&&(j+1)<len1&&dp[i][j+1]!=1){ 
            sum++; t=j; i++; t++;
            while(i<len2){                          
                while(t<len1&&dp[i][t]!=1) {
                    t++; sum++;
                }
                i++;  
            }    
            if(sum<sumt){
                sumt=sum;
                sum=0; 
                start=j;
                end=t;
            }
        }
    }
}
int main (){
    vector<string> data;
    string mm;
    while(cin>>mm){
        data.push_back(mm);    
    
    int len=data.size();

    for(int i=0;i<len;i=i+2){
        int x=0,y=0;
        cal(data[i],data[i+1],x,y);
        cout<<x<<" "<<y<<endl;
    }
}