Valid Anagram

class Solution {
public:
    bool isAnagram(string s, string t) {
        if(t=="") return s=="";
        map<char,int> m_s;
        map<char,int> m_t;
        int i=0;
        
        while(i<t.length()){
         if(m_t.find(t[i]) != m_t.end()) m_t[t[i]]++;
         else m_t.insert(pair<char,int>(t[i],1));
         i++;   
        }
        
        i=0;
        while(i<s.length()){
         if(m_s.find(s[i]) != m_s.end()) m_s[s[i]]++;
         else m_s.insert(pair<char,int>(s[i],1));
         i++;   
        }
        
         map<char,int >::iterator it;
         for(it=m_s.begin();it!=m_s.end();it++){   //s belong t
             if(m_t[it->first] != it->second)
                {return false;}
         }
         
         for(it=m_t.begin();it!=m_t.end();it++){  //t belong s
             if(m_s[it->first] != it->second)
                {return false;}
         }
         
        return true;
        
        
    }
};
原文地址:https://www.cnblogs.com/julie-yang/p/4907336.html