# 주어진 순열중에 1개만 존재하는 수 찾기 C++

# map 을 쓰지않는 방법으로 추후 구현해서 올릴 예정.

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#include <iostream> #include <vector> #include <map> using namespace std; int Solution(const vector<int>& numbers); void main() { vector<int> testVec; testVec.push_back(1); testVec.push_back(1); testVec.push_back(3); testVec.push_back(3); testVec.push_back(5); testVec.push_back(4); testVec.push_back(4); testVec.push_back(6); testVec.push_back(6); int answer = Solution(testVec); cout << "입력받은 숫자열중에 1개만 존재하는 숫자는 : " << answer << endl; system("pause"); } int Solution(const vector<int>& numbers) { int aloneNumber = 0; map<int, int> duplicateCheckMap; bool isInsertSuccess; for each (int num in numbers) { isInsertSuccess = duplicateCheckMap.insert(pair<int, int>(num, 1)).second; if (isInsertSuccess == false) { duplicateCheckMap[num] = 2; } } for each (pair<int, int> num in duplicateCheckMap) { if (num.second == 1) { aloneNumber = num.first; } } duplicateCheckMap.clear(); return aloneNumber; }