【Anagrams】 cpp

时间:2023-03-09 09:53:33
【Anagrams】 cpp

题目:

Given an array of strings, return all groups of strings that are anagrams.

Note: All inputs will be in lower-case.

代码:

class Solution {
public:
vector<string> anagrams(vector<string>& strs) {
vector<string> ret;
map<string,vector<int> > str_indexs;
// trans strs to key (sorted string) and value (indexs of strs that have same key) pairs
for ( size_t i = ; i < strs.size(); ++i )
{
string key = strs[i];
std::sort(key.begin(), key.end());
str_indexs[key].push_back(i);
}
// add the keys which occurs more than once
for ( map<string,vector<int> >::iterator it = str_indexs.begin(); it!= str_indexs.end(); ++it )
{
if ( it->second.size()> )
{
for ( size_t k = ; k < it->second.size(); ++k )
{
ret.push_back(strs[it->second[k]]);
}
}
}
return ret;
}
};

tips:

对于std::map的使用还不太熟悉,接着这道题也熟悉一下。

这道题非常经典,直接搜的AC code。学习了一下,把看过的几个blog记录在下面。

http://bangbingsyb.blogspot.sg/2014/11/leetcode-anagrams.html

http://www.cnblogs.com/AnnieKim/archive/2013/04/25/3041982.html

================================================

第二次过这道题,忘记了算法了。这道题做法确实比较特殊。先把每个单词按照字典序进行排序,然后直接把单词作为hash table的key进行判断。

最后输出重复count大于1的字符。注意排序的时候,不要直接拿strs进行排序,否则没法返回结果了。

class Solution {
public:
vector<string> anagrams(vector<string>& strs) {
map<string, vector<int> > string_count;
for ( int i=; i<strs.size(); ++i )
{
string tmp = strs[i];
std::sort(tmp.begin(), tmp.end());
string_count[tmp].push_back(i);
}
vector<string> ret;
for ( map<string, vector<int> >::iterator i=string_count.begin(); i!=string_count.end(); ++i )
{
if ( i->second.size()> )
{
for ( int j=; j<i->second.size(); ++j )
{
ret.push_back(strs[i->second[j]]);
}
}
}
return ret;
}
};

另外注意有个地方需要改进一下,最后往ret填充的时候,可以一个group一起insert了。

class Solution {
public:
vector<string> anagrams(vector<string>& strs) {
map<string, vector<string> > string_count;
for ( int i=; i<strs.size(); ++i )
{
string tmp = strs[i];
std::sort(tmp.begin(), tmp.end());
string_count[tmp].push_back(strs[i]);
}
vector<string> ret;
for ( map<string, vector<string> >::iterator i=string_count.begin(); i!=string_count.end(); ++i )
{
if ( i->second.size()> )
{
ret.insert(ret.end(), i->second.begin(), i->second.end());
}
}
return ret;
}
};