蔚来技术笔试AC

三道算法
1，2简单题
3、两个字符数S，T，每次可以将S中的一个移动到末尾，最少移动多少次使S==T
#include <iostream>
#include <iterator>
#include <map>
using namespace std;

int main() {
string S,T;
cin >> S >> T;
if(S==T)return 0;
int len = S.length();
int result =len;
map<char,int> S_map{{'a',0},{'b',0},{'c',0},{'d',0},{'e',0},{'f',0},{'g',0},{'h',0},{'i',0},{'j',0},{'k',0},{'l',0},{'m',0},{'n',0},{'o',0},{'p',0},{'q',0},{'r',0},{'s',0},{'t',0},{'u',0},{'v',0},{'w',0},{'x',0},{'y',0},{'z',0}};
map<char,int> T_map{{'a',0},{'b',0},{'c',0},{'d',0},{'e',0},{'f',0},{'g',0},{'h',0},{'i',0},{'j',0},{'k',0},{'l',0},{'m',0},{'n',0},{'o',0},{'p',0},{'q',0},{'r',0},{'s',0},{'t',0},{'u',0},{'v',0},{'w',0},{'x',0},{'y',0},{'z',0}};
for(int i=0;i<len;i++){
S_map[S[i]]+=1;
T_map[T[i]]+=1;
}
for(map<char,int>::iterator itr=S_map.begin();itr!=S_map.end();itr++){
if(T_map[itr->first]!=itr->second)return -1;
}
for(int i=0;i<len;i++){
int local=0;
local = S.find(T[i]);
if(local ==-1)break;
else{
S.erase(S.begin(),S.begin()+local+1);
result-=1;
}
}
cout<<result<<endl;
}