【程序员面试宝典】确定两串乱序同构

题目不难，构建两个map进行遍历查询就好

class Same { public: bool checkSam(string stringA, string stringB) { if(stringA.length()!=stringB.length()) { return false; } map<char,int> mapA; map<char,int> mapB; for(int i=0;i!=stringA.length();++i) { mapA[stringA[i]]++; mapB[stringB[i]]++; } for(auto it=mapA.begin();it!=mapA.end();++it) { if(it->second!=mapB[it->first]) { return false; } } return true; } };