Leetcode 49 Group Anagrams

class Solution {
public:
typedef map<char, int>::iterator MIT;
vector<vector<string>> groupAnagrams(vector<string>& strs) {
// mark the work
vector<vector<string> > ret;

vector<map<char,int> > maps;

size_t n = strs.size();

for (size_t i = 0; i < n; ++i) {
string str = strs[i];
size_t strlen = str.size();

map<char, int> tmp;
for (size_t j = 0; j < strlen; ++j) {
++tmp[str[j]];
}
size_t k = 0;
for (; k != maps.size(); ++k) {
if (maps[k].size() != tmp.size())  continue;

MIT itmp = tmp.begin();
for (; itmp != tmp.end(); ++itmp) {
MIT imaps = maps[k].find(itmp->first);
if (maps[k].end() == imaps || imaps->second != itmp->second)
break;
}

if (itmp != tmp.end())
continue;
else {
ret[k].push_back(str);
break;
}
}

// 没找到
if (k == maps.size()) {
ret.push_back(vector<string>(1, str));
maps.push_back(tmp);
}
}

for (size_t m = 0; m <ret.size(); ++m) {
sort(ret[m].begin(), ret[m].end());
}

return ret;
}
};

class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
unordered_map <string, multiset<string> > mp;

for (auto str : strs) {

// for (size_t i = 0; i != strs.size(); ++i) {
//string str = strs[i];
string keystr = sortStr(str);
// sort(keystr.begin(), keystr.end());
mp[keystr].insert(str);
}

vector<vector<string> > ret;
for (auto m : mp) {
ret.push_back(vector<string>(m.second.begin(), m.second.end()));
}
return ret;
}

private:
// sort 时间复杂度 O(n*logn)
// sortStr 时间复杂度 o(n)
string sortStr(string s) {

string ret;

vector<int> flag(26, 0);
for (auto ch : s) {
++flag[ch - 'a'];
}

for (size_t i = 0; i < 26; ++i) {
ret += string(flag[i], 'a' + i);
}
return ret;
}
};