Anagrams

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"]

 

anagrams是某个单词打乱其字母顺序形成的新单词，新单词和原单词包含的字母种类相同，每个字母的数目也相同。

因此可以使用哈希表存储单词的原型，没扫描一个单词，如果该单词原型在哈希表中，那么该单词就输出到结果中

class Solution {
public:
    typedef unordered_map<string, int> UMP;
    
public:
    vector<string> anagrams(vector<string> &strs) {
        UMP htable;
        vector<string> ans;
        for(int i=0; i<strs.size(); ++i) {
            string str = strs[i];
            sort(str.begin(), str.end());   //排序单词，认为其字典排序为其原型
            UMP::iterator iter = htable.find(str);  //在哈希表中查找其原型
            if( iter == htable.end() ) {    //找不着，就在哈希表中添加其原型
                htable.insert( make_pair(str, i) );
            }
            else {  //找到
                if( iter->second != -1 ) {  //如果结果中没有输出推导出原型的单词，则输出
                    ans.push_back(strs[iter->second]);
                    iter->second = -1;
                }
                ans.push_back(strs[i]); //哈希表中有原型，则输出该单词
            }
        }
        return ans;
    }
};

 

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        map<string, list<int>> mp_list;
        for (int i = 0; i < strs.size(); ++i) {
            string tmp = strs[i];
            sort(tmp.begin(), tmp.end());
            mp_list[tmp].push_back(i);
        }
        
        vector<vector<string>> ans;
        for (map<string, list<int>>::iterator iter = mp_list.begin();
             iter != mp_list.end();
             ++iter) {
            const list<int>& lt = iter->second;
            vector<string> strings;
            for (list<int>::const_iterator lt_iter = lt.begin();
                 lt_iter != lt.end();
                 ++lt_iter) {
                strings.push_back(strs[*lt_iter]);
            }
            sort(strings.begin(), strings.end());
            ans.push_back(strings);
        }
        
        return ans;
    }
};