【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 = 0; 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()>1 )
            {
                for ( size_t k = 0; 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=0; 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()>1 )
                {
                    for ( int j=0; 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=0; 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()>1 )
                {
                    ret.insert(ret.end(), i->second.begin(), i->second.end());
                }
            }
            return ret;
    }
};
原文地址:https://www.cnblogs.com/xbf9xbf/p/4495424.html