{"data":{"site":{"siteMetadata":{"title":"HOME | blackpudding","subtitle":"고생했어. 오늘도.","copyright":"© All rights reserved.","author":{"name":"William JO","github":"blackpudding96"},"disqusShortname":"blackpudding","url":"https://blackpudding.netlify.com"}},"markdownRemark":{"id":"7c7c3c4c-b755-5772-b3b4-40fcbdb2cd89","html":"

5430번 : AC

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👉 문제 바로가기

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배경 지식

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푼 방법

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R 함수와 D 함수를 입력 받은 순서 대로 진행해주고 나머지 원소들을 출력해주면 된다.

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여기서 중요한 포인트는 R 함수를 실행할 때 모든 원소의 순서를 뒤집으면 안된다. n이 커지고 R 함수가 많을수록 시간복잡도가 늘어나 time out(시간 초과) 날 수 있다.

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그러면 어떻게 이 문제를 해결해야 할까? 덱(deque)을 사용, R 함수를 실행할 때마다 flag 변수로 앞에서 삭제할지, 뒤에서 삭제할지 결정한다. 배열을 뒤집는 것은 배열의 뒷 부분을 앞으로 보는 것과 같기 때문에 덱(deque)의 특징이자 장점을 이용하여 앞, 뒤에서 자유롭게 원소를 삭제해나가면 된다.

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내 정답 코드 (C++)

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#include <iostream>\n#include <string>\n#include <sstream>\n\ntemplate <typename T>\nstruct Node {\n\tT data;\n\tstruct Node *front;\n\tstruct Node *rear;\n};\n\ntemplate <typename T>\nclass Deque {\nprivate:\n\tNode<T> *head;\n\tNode<T> *tail;\n\tint size;\n\t\npublic:\n\tDeque() : head(nullptr), tail(nullptr), size(0) {}\n\t~Deque() {\n\t\tif (this->head == nullptr)\n\t\t\treturn;\n\t\t\n\t\tNode<T> *cur = this->tail;\n\t\t\n\t\twhile (cur != nullptr) {\n\t\t\tcur = cur->front;\n\t\t\tdelete this->tail;\n\t\t\tthis->tail = cur;\n\t\t}\n\t\t\n\t\tthis->head = this->tail;\n\t}\n\t\n\tint get_size() {\n\t\treturn this->size;\n\t}\n\t\n\tbool is_empty() {\n\t\treturn this->head == nullptr ? true : false;\n\t}\n\t\n\tvoid push_front(T value) {\n\t\tNode<T> *node = new Node<T>;\n\t\tnode->data = value;\n\t\tnode->front = nullptr;\n\t\tnode->rear = nullptr;\n\t\t\n\t\tif (is_empty()) // first push.\n\t\t\tthis->tail = node;\n\t\telse {\n\t\t\tnode->rear = this->head;\n\t\t\tthis->head->front = node;\n\t\t}\n\t\t\n\t\tthis->head = node;\n\t\t++size;\n\t}\n\t\n\tvoid push_back(T value) {\n\t\tNode<T> *node = new Node<T>;\n\t\tnode->data = value;\n\t\tnode->front = nullptr;\n\t\tnode->rear = nullptr;\n\t\t\n\t\tif (is_empty()) // first push.\n\t\t\tthis->head = node;\n\t\telse {\n\t\t\tnode->front = this->tail;\n\t\t\tthis->tail->rear = node;\n\t\t}\n\t\t\n\t\tthis->tail = node;\n\t\t++size;\n\t}\n\t\n\tT pop_front() {\n\t\tif (is_empty()) // empty deque.\n\t\t\treturn -1; // T type에 맞는 error code return.\n\t\telse {\n\t\t\tint pop_val = this->head->data;\n\t\t\t\n\t\t\tif (this->head == this->tail) { // deque with 1 element.\n\t\t\t\tdelete this->head;\n\t\t\t\tthis->head = this->tail = nullptr;\n\t\t\t}\n\t\t\telse { // deque with more than 2 elements.\n\t\t\t\tNode<T> *node = new Node<T>;\n\t\t\t\t\n\t\t\t\tnode = this->head->rear;\n\t\t\t\t\n\t\t\t\tdelete this->head;\n\t\t\t\t\n\t\t\t\tthis->head = node;\n\t\t\t\tthis->head->front = nullptr;\n\t\t\t}\n\t\t\t\n\t\t\t--size;\n\t\t\treturn pop_val;\n\t\t}\n\t}\n\t\n\tT pop_back() {\n\t\tif (is_empty()) // empty deque.\n\t\t\treturn -1; // T type에 맞는 error code return.\n\t\telse {\n\t\t\tint pop_val = this->tail->data;\n\t\t\t\n\t\t\tif (this->head == this->tail) { // deque with 1 element.\n\t\t\t\tdelete this->head;\n\t\t\t\tthis->head = this->tail = nullptr;\n\t\t\t}\n\t\t\telse { // deque with more than 2 elements.\n\t\t\t\tNode<T> *node = new Node<T>;\n\t\t\t\t\n\t\t\t\tnode = this->tail->front;\n\t\t\t\t\n\t\t\t\tdelete this->tail;\n\t\t\t\t\n\t\t\t\tthis->tail = node;\n\t\t\t\tthis->tail->rear = nullptr;\n\t\t\t}\n\t\t\t\n\t\t\t--size;\n\t\t\treturn pop_val;\n\t\t}\n\t}\n};\n\nint main() {\n\tstd::ios::sync_with_stdio(false);\n\tstd::cin.tie(NULL); \n\tstd::cout.tie(NULL);\n\t\n\tbool from_front, error;\n\tDeque<int> deque;\n\tint T, n;\n\tstd::string p, num_arr;\n\n\tstd::cin >> T;\n\t\n\tfor (int i = 0; i < T; ++i) {\n\t\tfrom_front = true, error = false;\n\t\tstd::cin >> p;\n\t\tstd::cin >> n;\n\t\tstd::cin >> num_arr;\n\t\t\n\t\tnum_arr = num_arr.substr(1); // '[' 제거\n\t\tnum_arr = num_arr.substr(0, num_arr.size() - 1); // ']' 제거\n\t\t\n\t\tif (n > 0) {\n\t\t\tstd::string token;\n\t\t\tstd::stringstream ss(num_arr);\n\t\t\t\n\t\t\twhile (std::getline(ss, token, ','))\n\t\t\t\tdeque.push_back(atoi(token.c_str()));\n\t\t}\n\t\t\n\t\t// R 함수 && D 함수 진행\n\t\tfor (int j = 0; j < p.size(); ++j) {\n\t\t\tif (p[j] == 'R') {\n\t\t\t\tfrom_front = !from_front;\n\t\t\t}\n\t\t\telse if (p[j] == 'D') {\n\t\t\t\tif (from_front) {\n\t\t\t\t\tif (deque.pop_front() == -1)\n\t\t\t\t\t\terror = true;\n\t\t\t\t}\n\t\t\t\telse {\n\t\t\t\t\tif(deque.pop_back() == -1)\n\t\t\t\t\t\terror = true;\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t\t\n\t\t// 결과 출력\n\t\tif (error)\n\t\t\tstd::cout << \"error\" << '\\n';\n\t\telse {\n\t\t\tif (deque.is_empty())\n\t\t\t\tstd::cout << \"[]\" << '\\n';\n\t\t\telse {\n\t\t\t\tstd::cout << \"[\";\n\t\t\t\t\n\t\t\t\tif (from_front) {\n\t\t\t\t\twhile (deque.get_size() >= 2)\n\t\t\t\t\t\tstd::cout << deque.pop_front() << \",\";\n\t\t\t\t\tstd::cout << deque.pop_front() << \"]\\n\";\n\t\t\t\t}\n\t\t\t\telse {\n\t\t\t\t\twhile (deque.get_size() >= 2)\n\t\t\t\t\t\tstd::cout << deque.pop_back() << \",\";\n\t\t\t\t\tstd::cout << deque.pop_back() << \"]\\n\";\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t}\n\t\n\treturn 0;\n}
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References

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https://psychoria.tistory.com/666
\nhttps://geekhub.tistory.com/63

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