LeetCode – Refresh – Anagrams

Use a hash table to record it. The tricky part is that when the hash value is > 0, it is the index + 1. Otherwise, that string already exists in result list.

 

class Solution {
public:
    vector<string> anagrams(vector<string> &strs) {
        vector<string> result;
        int len = strs.size();
        if (len < 2) return result;
        unordered_map<string, int> mapping;
        for (int i = 0; i < len; i++) {
            string tmp = strs[i];
            sort(tmp.begin(), tmp.end());
            if (mapping[tmp]) {
                if (mapping[tmp] > 0) {
                    result.push_back(strs[mapping[tmp]-1]);
                    mapping[tmp] = -1;
                }
                result.push_back(strs[i]);
            } else {
                mapping[tmp] = i+1;
            }
        }
        return result;
    }
};