Notes: | |
Given an array of strings, return all groups of strings that are anagrams. | |
Note: All inputs will be in lower-case. | |
Solution: Sort the string to see if they're anagrams. | |
Solution 1 is simpler than 2. | |
*/ |
class Solution {
public:
vector<string> anagrams(vector<string> &strs) {
return anagrams_1(strs);
}
// solution 1
vector<string> anagrams_1(vector<string> &strs) {
typedef map<string, vector<int> > MAP;
MAP map;
for (int i = 0; i < strs.size(); ++i)
{
string s = strs[i];
sort(s.begin(), s.end());
map[s].push_back(i);
}
vector<string> res;
MAP::iterator it = map.begin();
for (; it != map.end(); it++)
{
vector<int> &anagrams = it->second;
if (anagrams.size() > 1) {
for (int i = 0; i < anagrams.size(); ++i)
res.push_back(strs[anagrams[i]]);
}
}
return res;
}
// solution 2
vector<string> anagrams_2(vector<string> &strs) {
typedef unordered_map<string, int > MAP;
vector<string> res;
MAP anagram;
for (int i = 0; i < strs.size(); ++i)
{
string s = strs[i];
sort(s.begin(), s.end());
MAP::iterator it = anagram.find(s);
if (it == anagram.end())
{
anagram[s] = i;
}
else
{
if (it->second >= 0) {
res.push_back(strs[it->second]);
it->second = -1;
}
res.push_back(strs[i]);
}
}
return res;
}
};