Anagrams

题目：

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

Note: All inputs will be in lower-case.

class Node{
public:
	vector<char> bit;
	int len;
	int index;
	int   operator <  (const Node& right) const {
		if(this->len != right.len){
			return this->len < right.len;
		} else {
			
			for(int i = 0; i < 26; i++){

				if(bit[i] != right.bit[i]){
					return bit[i] < right.bit[i];
				}
			}
			return false;
		}
	}
	
	int   operator ==  (const Node& right) const {
		if(this->len != right.len){
			return false;
		} else {
			if(this->len == 0){
				return true;
			}
			for(int i = 0; i < 26; i++){

				if(bit[i] != right.bit[i]){
					return false;
				}
			}
			return true;
		}
	}
};
class Solution {
public:
	
    vector<string> anagrams(vector<string> &strs) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int len = strs.size();
		vector<string> ans;
		if(len < 2){
			return ans;
		}
		
		vector<Node> vec;
		
		int i, j;
		for(i = 0; i < len; i++){
			Node node;
			vector<char> a(32, 0);
			node.len = strs[i].size();
			if(node.len != 0){
				for(j = 0; j < node.len; j++){
					a[strs[i][j] - 'a']++;
				}
			}
			node.bit = a;
			node.index = i;
			vec.push_back(node);
		}
		sort(vec.begin(), vec.end(), less<Node>());
		for(int i = 0; i< len - 1; i++){
			if(vec[i] == vec[i+1]){
				ans.push_back(strs[vec[i].index]);
				ans.push_back(strs[vec[i + 1].index]);
				int j = 2;
				while(i + j < len && vec[i] == vec[i+j]){
					ans.push_back(strs[vec[i + j].index]);
					j++;
				}
				i = i+j - 1;
			}
		}
		return ans;
    }
};