LeetCode Anagrams

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

Note: All inputs will be in lower-case.

这题最开始的思路是用map<string, vector<sting> >来保存与string对应的anagrams，然后有一张vector<string> word来保存key string，但问题是每次新的string要遍历word来进行比较是否存在，用的是一个count[26]来保存字符的个数，然后一一比较。

最后发现其实先把每个string按字母排个序，然后就能很方便的进行string的两两比较了。写了3个版本。

这题的关键是对每个string按字典对其char重新排序，这样就能方便的得出map<string, vector<string> >中的key了

把sort string作为key插入set来比较

class Solution {
private:
    vector<string> ret;
    vector<bool> canUse;
    set<string> s;
    map<string, int> index;
public:    
    vector<string> anagrams(vector<string> &strs) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ret.clear();
        s.clear();
        
        vector<string> sortStr(strs);
        
        canUse.resize(strs.size());
        
        for(int i = 0; i < canUse.size(); i++)
            canUse[i] = false;
        
        for(int i = 0; i < sortStr.size(); i++)
            sort(sortStr[i].begin(), sortStr[i].end());
            
        for(int i = 0; i < sortStr.size(); i++)
            if (s.count(sortStr[i]) > 0)
            {
                ret.push_back(strs[i]);
                canUse[index[sortStr[i]]] = true;
            }
            else
            {
                index[sortStr[i]] = i;
                s.insert(sortStr[i]);
            }
            
        
        for(int i = 0; i < canUse.size(); i++)
            if (canUse[i])
                ret.push_back(strs[index[sortStr[i]]]);    
        
        return ret;
    }
};

sort string后，根据sort string来排序，找出同类的string

struct Node
{
    string org;
    string sortOrg;
    Node(){}
    Node(string &a, string &b):org(a), sortOrg(b){}
};

bool comp(const Node &lhs, const Node &rhs)
{
    return lhs.sortOrg < rhs.sortOrg;
}

class Solution {
private:
    vector<Node> a;
    vector<string> ret;
public:
    vector<string> anagrams(vector<string> &strs) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ret.clear();
        a.clear();
        if (strs.size() == 0)
            return ret;
            
        for(int i = 0; i < strs.size(); i++)
        {
            string sortStrs(strs[i]);
            sort(sortStrs.begin(), sortStrs.end());
            
            a.push_back(Node(strs[i], sortStrs));
        }
        
        sort(a.begin(), a.end(), comp);
        
        string key = "1";
        
        int start = -1;
        int end = 0;
        
        while(end < a.size())
        {
            if (key == a[end].sortOrg)
                end++;
            else
            {
                if (start + 1 != end)
                {
                    for(int i = start; i < end; i++)
                        ret.push_back(a[i].org);
                }
                key = a[end].sortOrg;
                start = end;
                end++;
            }
        }
        
        if (start + 1 != end)
        {
            for(int i = start; i < end; i++)
                ret.push_back(a[i].org);
        }
        
        return ret;
    }
};

不用保存word，直接sort string后得出key

class Solution {
private:
    vector<string> ret;
    map<string, vector<string> > m;
public:
    vector<string> anagrams(vector<string> &strs) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ret.clear();
        m.clear();
        for(int i = 0; i < strs.size(); i++)
        {
            string sortStr(strs[i]);
            sort(sortStr.begin(), sortStr.end());
            
            m[sortStr].push_back(strs[i]);
        }
        
        for(map<string, vector<string> >::iterator iter = m.begin(); iter != m.end(); iter++)
        {
            if ((iter->second).size() > 1)
            {
                for(int i = 0; i < (iter->second).size(); i++)
                    ret.push_back((iter->second)[i]);
            }
        }
        
        return ret;
    }
};