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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.
1 class Solution { 2 public: 3 vector<string> anagrams(vector<string> &strs) { 4 vector<string> result; 5 int len = strs.size(); 6 if (len < 2) return result; 7 unordered_map<string, int> mapping; 8 for (int i = 0; i < len; i++) { 9 string tmp = strs[i]; 10 sort(tmp.begin(), tmp.end()); 11 if (mapping[tmp]) { 12 if (mapping[tmp] > 0) { 13 result.push_back(strs[mapping[tmp]-1]); 14 mapping[tmp] = -1; 15 } 16 result.push_back(strs[i]); 17 } else { 18 mapping[tmp] = i+1; 19 } 20 } 21 return result; 22 } 23 };
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原文地址:http://www.cnblogs.com/shuashuashua/p/4346158.html