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.

For example:

Input:　　["tea","and","ate","eat","den"]

Output: ["tea","ate","eat"]

Interface: vector<string>anagrams(vector<string>&strs);

#include <iostream>#include <vector>#include <unordered_map>#include <string>#include <algorithm>using namespace std;//方法一//用unordered_map<string,vector<string>> anagram 记录排序后的字符串与未排序的字符串//如果对应的未排序字符串数量超过一个，记录到新的vector中vector<string> anagrams(vector<string> &strs){vector<string> res;string s;if(strs.size() <= 1) return res;unordered_map<string,vector<string>> anagram;for(int i=0;i<strs.size();i++){s = strs[i];sort(s.begin(),s.end());anagram[s].push_back(strs[i]);}for(auto it=anagram.begin();it!=anagram.end();it++){if(it->second.size()>1){res.insert(res.end(),it->second.begin(),it->second.end());}}return res;}//方法二//思路：用map<string, int>记录排序后的字符串以及首次出现的位置。//1. 从strs的第一个元素开始遍历，首先对元素进行排序得到s；//2. 在map里查找s；//3. 若不存在，将s以及该元素的下标存入map<string ,int>；//4. 若存在，首先将第一次出现s时的原始字符串存入结果res，即strs[map[s]]，并将map[s]设置为-1（防止下次再存），再将该字符串本身存入结果res；//5. 重复以上1-4步，直到遍历结束。vector<string> anagrams(vector<string> &strs){vector<string> res;string s;if(strs.size() <= 1) return res;unordered_map<string,int> anagram;for(int i=0;i<strs.size();i++){s = strs[i];sort(s.begin(),s.end());if (anagram.find(s) == anagram.end())anagram.insert(pair<string,int>(s,i));else{if(anagram[s]>=0){res.push_back(strs[anagram[s]]);anagram[s] = -1;}res.push_back(strs[i]);}}return res; }void showVector(vector<string> &strs){for(auto it=strs.begin();it!=strs.end();it++){printf("%s ",(*it).c_str());}//for(int i =0;i<strs.size();i++)//{//printf("%s ",strs[i].c_str());//}printf("\n");}void main(){vector <string> res;string array[] = {"abc","bac","cab","123","efg"};int len = sizeof(array)/sizeof(string);vector <string> strs(array,array+len);showVector(strs);res = anagrams(strs);showVector(res);}

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