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 #include #include using namespace std;int main() {string S,T;cin >> S >> T;if(S==T)return 0;int len = S.length();int result =len;map 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 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;iS_map[S[i]]+=1;T_map[T[i]]+=1;}for(map::iterator itr=S_map.begin();itr!=S_map.end();itr++){if(T_map[itr->first]!=itr->second)return -1;}for(int i=0;iint local=0;local = S.find(T[i]);if(local ==-1)break;else{S.erase(S.begin(),S.begin()+local+1);result-=1;}}cout}
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