【数据结构_map容器迭代】Anagrams

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Anagrams

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Given an array of strings, return all groups of strings that are anagrams.

Note: All inputs will be in lower-case.

关键是string的内部字符排序和map容器的迭代

class Solution {public:    vector<string> anagrams(vector<string> &strs) {        vector<string> res;int n=strs.size();        if(n==0) return res;                map<string,vector<int> > ss;        char buffer[1000];        for(int i=0;i<n;i++){            strcpy(buffer,strs[i].c_str());//从string拷贝到char*            sort(buffer,buffer+strs[i].length());//并排序char*            string s=buffer;                       //再转成string            ss[s].push_back(i);        }        for( map<string,vector<int> >::iterator it=ss.begin();it!=ss.end();it++){//map迭代，*it是pair<?,?>            vector<int> &t=it->second;            if(t.size()>1){                for(int i=0;i<t.size();i++)                     res.push_back(strs[t[i]]);            }        }        return res;    }};

或者对pair <sorted_string,string> [ ]进行排序,比较pair.first以查重。

class Solution {public:    vector<string> anagrams(vector<string> &strs) {        vector<pair<string,string> > vec;vector<string> ans;        typedef pair<string,string> Pair;        int n=strs.size();        if(n==0) return ans;                for(int i=0;i<n;i++){            string s=strs[i];            sort(s.begin(),s.end());            vec.push_back(Pair(s,strs[i]));        }        sort(vec.begin(),vec.end());                string last=vec[0].first;int m=0;        for(int i=0;i<=vec.size();i++){            if(i<vec.size()&&vec[i].first==last) m++;            else{                if(m>1){                    for(int j=i-1;j>=i-m;j--) ans.push_back(vec[j].second);                }                if(i<vec.size()){                    last=vec[i].first,m=1;                }            }        }        return ans;    }};

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