leetcode@ [49] Group Anagrams (Hashtable)

https://leetcode.com/problems/anagrams/

Given an array of strings, group anagrams together.

For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"],
Return:

[
  ["ate", "eat","tea"],
  ["nat","tan"],
  ["bat"]
]

 

Note:

1. For the return value, each inner list's elements must follow the lexicographic order.
2. All inputs will be in lower-case.

class node {
    public:  
        int idx;
        string s;
        node(int i, string str) {
            idx =i;
            s = str;
        }
        bool operator < (const node& rhs) {
            if(s.compare(rhs.s) < 0 || (s.compare(rhs.s) == 0 && idx < rhs.idx))  return true;
            return false;
        }
};

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        vector<vector<string> > res;
        
        int n = strs.size();
        if(n == 0)  return res;
        if(n == 1) {
            vector<string> tmp;
            tmp.push_back(strs[0]);
            res.push_back(tmp);
            return res;
        }
        
        vector<vector<char> > vec(n);
        vector<node> cs;
        for(int i=0; i<n; ++i) {
            for(int j=0; j<strs[i].length(); ++j) {
                vec[i].push_back(strs[i][j]);
            }
            sort(vec[i].begin(), vec[i].end());
            
            string ss = "";
            for(int k=0; k<vec[i].size(); ++k) {
                ss += vec[i][k];
            }
            cs.push_back(node(i, ss));
        }
        sort(cs.begin(), cs.end());
        
        int l = 0, r = l+1;
        while(r < n) {
            while(r < n && (cs[l].s).compare(cs[r].s) == 0)  ++r;
            
            vector<string> row;
            for(int p=l; p<r; ++p)  row.push_back(strs[cs[p].idx]);
            res.push_back(row);
            
            l = r; r = l+1;
        }
        
        if(l < n) {
            vector<string> row;
            row.push_back(strs[cs[n-1].idx]);
            res.push_back(row);
        }
        for(int i=0; i<res.size(); ++i) {
            sort(res[i].begin(), res[i].end());
        }
        
        return res;
    }
};