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"]
所谓的anagrams，就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，每个字母的数目相同。 本文地址
用哈希map存储排序后的字符串，map中key为排序后字符串，value为该字符串对应的第一个原字符串在数组中的位置。如果value = -1，表示该字符串对应的第一个源字符串已经输出过
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class
Solution {
public
:
vector<string> anagrams(vector<string> &strs) {
typedef
unordered_map<string,
int
> Umap;
Umap hashtable;
vector<string> res;
for
(
int
i = 0; i < strs.size(); i++)
{
string s = strs[i];
sort(s.begin(), s.end());
Umap::iterator ite = hashtable.find(s);
if
(ite == hashtable.end())
hashtable.insert(Umap::value_type(s, i));
else
{
if
(ite->second != -1)
{
res.push_back(strs[ite->second]);
ite->second = -1;
}
res.push_back(strs[i]);
}
}
return
res;
}
};
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本文转自tenos博客园博客，原文链接：http://www.cnblogs.com/TenosDoIt/p/3681402.html，如需转载请自行联系原作者
