{"componentChunkName":"component---src-templates-blog-post-js","path":"/Interview/[면접준비]-알고리즘/","result":{"data":{"site":{"siteMetadata":{"title":"Hun's Footsteps 🥷","author":"전여훈","siteUrl":"https://jeonyeohun.netlify.app","comment":{"disqusShortName":"","utterances":"jeonyeohun/jeonyeohun.github.io"},"sponsor":{"buyMeACoffeeId":"jeonyeohun"}}},"markdownRemark":{"id":"070e4961-b943-58f9-b4be-877148a4d205","excerpt":"개발자 면접질문과 답변들. 더 많은 내용은 깃허브 저장소에서 확인할 수 있습니다 (https://github.com/jeonyeohun/GetReadyForInterview) 정렬 알고리즘 버블정렬 - O(N^2) 로직(오름차순) 처음부터 n - i 까지 인접한 인덱스를 비교해서 값이 가장 큰 요소를 n - i 번째에 위치시킵니다. i는 정렬이 완료된 요소의 개수를 의미합니다. 시간복잡도 N 개의 인덱스에 대해 최악의 경우 N 번 비교연산을 수행해야하기 때문에 O(N^…","html":"
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개발자 면접질문과 답변들. 더 많은 내용은 깃허브 저장소에서 확인할 수 있습니다 (https://github.com/jeonyeohun/GetReadyForInterview)

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정렬 알고리즘

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버블정렬 - O(N^2)

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void bubble_sort(vector<int> & list){\n  for (int i = 0 ; i < list.size() ; i++){\n    for (int j = 0 ; j < list.size() - i - 1 ; j++){\n      if(list[j] > list[j + 1]){\n        swap (list[j], list[j + 1]);\n      }\n    }\n  }\n}
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선택정렬 - O(N^2)

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void selection_sort (vector<int> & list){\n  for (int i = 0 ; i < list.size() ; i++){\n    for (int j = 0 ; j < list.size() ; j++){\n      if (list[i] < list[j]){\n        swap(list[i], list[j]);\n      }\n    }\n  }\n}
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삽입정렬 - O(N^2)

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void insertion_sort (vector<int> & list){\n  int i, j;\n\tfor (i = 1 ; i < list.size() ; i++){\n    int current = list[i];\n    for (j = i - 1 ; j >= 0 ; j--){\n      if(list[j] > current){\n        list[j + 1] = list[j];\n      }\n      else{\n        break;\n      }\n    }\n\n    list[j + 1] = current;\n  }\n}
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병합정렬 - O(NlogN)

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void merge (vector<int> & list, int left, int mid, int right){\n  vector<int> sorted_list(list.size());\n  int i = left, j = mid+1, idx = left;\n  while(i <= mid && j <= right){\n    if (list[i] <= list[j]){\n      sorted_list[idx++] = list[i++];\n    }\n    else{\n      sorted_list[idx++] = list[j++];\n    }\n  }\n  while(i <= mid){\n    sorted_list[idx++] = list[i++];\n  }\n\n  while(j <= right){\n    sorted_list[idx++] = list[j++];\n  }\n\n  for (int l = left ; l <= right ; l++){\n    list[l] = sorted_list[l];\n  }\n}\n\nvoid partition (vector<int> & list, int left, int right){\n  int mid;\n  if (left < right){\n    mid = (left + right) / 2;\n    partition(list, left, mid);\n    partition(list, mid+1, right);\n    merge(list, left, mid, right);\n  }\n}
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퀵 정렬 - O(NlogN)

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void quick_sort (vector<int> & list, int head, int tail){\n  if (head >= tail) return;\n\n  int pivot = head;\n  int left = pivot + 1;\n  int right = tail;\n\n  while(left <= right){\n    while(left <= tail && list[left] <= list[pivot]){\n      left++;\n    }\n    while(right > head && list[right] >= list[pivot]){\n      right--;\n    }\n\n    if (left > right){\n      swap(list[pivot], list[right]);\n    }\n    else{\n      swap(list[right], list[left]);\n    }\n  }\n\n  quick_sort(list, head, right - 1);\n  quick_sort(list, right + 1, tail);\n}
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피보나치 수열 구현

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재귀

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int fibonacci (int n){\n if (n < 2){\n   return n;\n }\n\n  return fibonacci(n - 1) + fibonacci(n - 2);\n}
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메모이제이션

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int memo[100] = {0,};\nint fibonacci (int n){\n if (n < 2){\n   return n;\n }\n if (memo[n]) return memo[n];\n return memo[n] = fibonacci(n - 1) + fibonacci(n - 2);\n}
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다이나믹 프로그래밍 - Bottom Up

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int fibonacci (int n){\n  int dp[n];\n  dp[1] = 1, dp[2] = 1;\n\n  for (int i = 3 ; i <= n ; i++){\n    dp[i] = dp[i - 1] + dp[i - 2];\n  }\n\n  return dp[n];\n}
","frontmatter":{"title":"[면접준비] 알고리즘","date":"May 24, 2021"}}},"pageContext":{"slug":"/Interview/[면접준비]-알고리즘/","previous":{"fields":{"slug":"/Interview/[면접준비]-웹/"},"frontmatter":{"title":"[면접준비] 웹","category":"Interview","draft":false}},"next":{"fields":{"slug":"/Interview/[면접준비]-운영체제/"},"frontmatter":{"title":"[면접준비] 운영체제","category":"Interview","draft":false}}}},"staticQueryHashes":["2486386679","3128451518"]}