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찬민 책 구현 숙제 (#70)
* 찬민 백준 숙제 * 찬민 구현숙제 * 찬민 코드
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‎cmkim/2020 카카오 인턴쉽/경주로 건설.py‎

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# 결국 그냥 미로찾기 문제에서 기본 비용은 100원 처리, 코너가 발생할때마다 500원 추가로 처리하면됀다
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from collections import deque
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board = [[0, 0, 1, 0, 1, 1, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 1, 1, 0, 1], [1, 0, 0, 0, 0, 1, 1, 0, 1, 0], [0, 0, 0, 0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 0, 1, 0, 1, 0, 1, 1], [0, 0, 1, 0, 1, 1, 0, 1, 0, 1], [0, 1, 0, 0, 1, 0, 0, 0, 1, 0], [1, 0, 0, 1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 0, 1, 0, 0], [1, 0, 0, 0, 0, 0, 0, 0, 1, 0]]
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q = deque()
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dx, dy = [-1, 1, 0, 0], [0, 0, -1, 1] #상하좌우
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INF = 1000000000
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def solution(board):
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answer = 0
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size = len(board)
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dp = [[INF] * size for _ in range(size)]
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visited = [[0] * size for _ in range(size)]
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visited[0][0] = 1
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if board[0][1] == 0:
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q.append((0, 0, 0, 1))
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while q:
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x, y, count, dir = q.popleft() #dir은 진행방향 1 = 가로, -1 = 세로
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# if x == size - 1 and y == size - 1:
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# break
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for i in range(4):
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nx = x + dx[i]
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ny = y + dy[i]
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if 0 <= nx < size and 0 <= ny < size and board[nx][ny] == 0:
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if dp[nx][ny] > count + 6:
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if dir == 1 and i % 4 < 2: # 진행방향이 가로고, 다음 움직일 곳이 세로면
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dp[nx][ny] = min(dp[nx][ny], count + 6)
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q.append((nx, ny, count + 6, -1))
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elif dir == -1 and i % 4 > 1: # 진행방향이 세로고, 다음 움직일 곳이 가로면
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dp[nx][ny] = min(dp[nx][ny], count + 6)
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q.append((nx, ny, count + 6, 1))
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if dp[nx][ny] > count + 1:
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if dir == 1 and i % 4 > 1: # 진행방향이 가로고, 다음 움직일 곳도 가로면
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dp[nx][ny] = min(dp[nx][ny], count + 1)
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q.append((nx, ny, count + 1, 1))
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elif dir == -1 and i % 4 < 2: # 진행방향이 세로고, 다음 움직일 곳이 세로면
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dp[nx][ny] = min(dp[nx][ny], count + 1)
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q.append((nx, ny, count + 1, -1))
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answer = dp[size - 1][size - 1] * 100
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if board[1][0] == 0:
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q.append((0, 0, 0, -1))
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visited = [[INF] * size for _ in range(size)]
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visited[0][0] = 1
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dp = [[INF] * size for _ in range(size)]
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while q:
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x, y, count, dir = q.popleft() # dir은 진행방향 1 = 가로, -1 = 세로
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# if x == size - 1 and y == size - 1:
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# break
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for i in range(4):
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nx = x + dx[i]
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ny = y + dy[i]
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if 0 <= nx < size and 0 <= ny < size and board[nx][ny] == 0:
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if dp[nx][ny] > count + 6:
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if dir == 1 and i % 4 < 2: # 진행방향이 가로고, 다음 움직일 곳이 세로면
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dp[nx][ny] = min(dp[nx][ny], count + 6)
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q.append((nx, ny, count + 6, -1))
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elif dir == -1 and i % 4 > 1: # 진행방향이 세로고, 다음 움직일 곳이 가로면
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dp[nx][ny] = min(dp[nx][ny], count + 6)
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q.append((nx, ny, count + 6, 1))
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if dp[nx][ny] > count + 1:
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if dir == 1 and i % 4 > 1: # 진행방향이 가로고, 다음 움직일 곳도 가로면
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dp[nx][ny] = min(dp[nx][ny], count + 1)
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q.append((nx, ny, count + 1, 1))
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elif dir == -1 and i % 4 < 2: # 진행방향이 세로고, 다음 움직일 곳이 세로면
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dp[nx][ny] = min(dp[nx][ny], count + 1)
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q.append((nx, ny, count + 1, -1))
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answer = min(answer, dp[size-1][size-1] * 100)
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return answer
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print(solution(board))

‎cmkim/2020 카카오 인턴쉽/보석 쇼핑.py‎

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def solution(gems):
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answer = []
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bosuck = []
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for i in gems:
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if i not in bosuck:
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bosuck.append(i)
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gems_count = len(gems)
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bosuck_count = len(bosuck)
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print(gems_count)
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switch = ([0] * bosuck_count for _ in range(gems_count))
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for i in range(gems_count):
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if gems[i] in bosuck:
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for j in range(bosuck_count):
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if gems[i] == bosuck[j]:
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switch[i][j] = 1
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print(switch)
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return answer
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'''
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[1] [0] [0] [0]
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[0] [1] [0] [0]
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[0] [0] [1] [0]
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[0] [0] [0] [1]
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[0] [0] [1] [0]
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[0] [0] [0] [0]
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[0] [0] [0] [0]
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[0] [0] [0] [0]
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'''

‎cmkim/2020 카카오 인턴쉽/수식 최대화.py‎

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# n가지 연산문자의 우선순위 재정의
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# 무조건 n가지 우선순위가 있어야한다
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# 음수는 절댓값으로 바꾸어 양수로 제출
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def solution(expression):
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answer = 0
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arr = '' #연산자 저장하는곳
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arr2 = ''
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num = []
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for i in range(len(expression)):
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if expression[i] == '+' or expression[i] == '-' or expression[i] == '*':
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#arr.append[i]
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arr += expression[i]
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#expression[i] = 0
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print(expression)
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# expression.replace('+', '.')
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# expression.replace('-', '.')
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# expression.replace('*', '.')
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table = str.maketrans('+-*', '...') #씨발련
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expression.translate(table)
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print(expression)
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arr2 = (expression.split('.'))
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print(arr2)
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return answer

‎cmkim/2020 카카��� 인턴쉽/키패드 누르기.py‎

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# 2, 5, 8, 0 의 경우 가장 최근의 왼손 오른손 위치 따지기 + 주손 따지기
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def solution(numbers, hand):
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answer = ''
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left_cur = 0 # 양손의 위치 | 1, 4, 7, * 순으로 3, 2, 1, 0 에 대응한다
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left_hor = 1 # hor 값은 가로로 1, 2, 3 번째 줄 중 값중 하나
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right_cur = 0
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right_hor = 3
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mid_cur = 0
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mid_hor = 2
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for i in numbers:
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if i == 1 or i == 4 or i == 7 or i == '*':
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#print(left_cur, left_hor, right_cur, right_hor, mid_cur, mid_hor)
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answer += 'L'
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if i == 1:
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left_cur = 3
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if i == 4:
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left_cur = 2
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if i == 7:
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left_cur = 1
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if i == '*':
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left_cur = 0
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left_hor = 1
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if i == 3 or i == 6 or i == 9 or i == '#':
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#print(left_cur, left_hor, right_cur, right_hor, mid_cur, mid_hor)
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answer += 'R'
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if i == 3:
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right_cur = 3
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if i == 6:
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right_cur = 2
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if i == 9:
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right_cur = 1
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if i == '#':
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right_cur = 0
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right_hor = 3
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if i == 2 or i == 5 or i == 8 or i == 0:
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#print(left_cur, left_hor, right_cur, right_hor, mid_cur, mid_hor)
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if i == 2:
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mid_cur = 3
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if i == 5:
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mid_cur = 2
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if i == 8:
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mid_cur = 1
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if i == 0:
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mid_cur = 0
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mid_hor = 2
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left_d = abs(mid_cur - left_cur) + (mid_hor - left_hor)
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right_d = abs(mid_cur - right_cur) + (right_hor - mid_hor) #abs가 절댓값이였나?
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if left_d < right_d:
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answer += 'L'
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left_cur = mid_cur
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left_hor = 2
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elif left_d > right_d:
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answer += 'R'
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right_cur = mid_cur
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right_hor = 2
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else:
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if hand == 'left':
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answer += 'L'
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left_cur = mid_cur
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left_hor = 2
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else:
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answer += 'R'
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right_cur = mid_cur
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right_hor = 2
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return answer

‎cmkim/Programmers/스택_큐/다리를 지나는 트럭.py‎

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def solution(bridge_length, weight, truck_weights):
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answer = 0
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return answer
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'''
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트럭 무게를 정렬할 필요는 없고
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그냥 순서대로
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'''
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‎cmkim/Programmers/정렬/H-index.py‎

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def solution(citations):
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answer = 0
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count = 0
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length = len(citations)
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citations.sort()
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# print(citations[-4])
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for i in range(1, length):
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if citations[-i] > i:
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answer = i
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return answer
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# 끝에서 N번째 수가 N보다 큰가? 에 대한 질문
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'''
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0 1 3 3 5 6
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1 2 2 3 3 4 4 4 5 6 7 9
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'''

‎cmkim/Programmers/정렬/K번째 수.py‎

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def solution(array, commands):
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arr = []
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answer = []
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example = len(commands)
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for i in range(example):
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start = commands[i][0]
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end = commands[i][1]
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count = commands[i][2]
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arr = array[start-1:end]
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arr.sort()
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#print(arr)
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answer.append(arr[count-1])
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return answer

‎cmkim/Programmers/정렬/가장 큰 수.py‎

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def solution(numbers):
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answer = ''
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count = 0
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for i in range(len(numbers)):
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count += len(str(numbers[i])) # i번째 숫자의 자리수를 더해줘서 총 자리수 구하기
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return answer
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9
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---------
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7.51
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7.53
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753
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755
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7.92
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75 / 10 = 7.55
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7 7.77
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7 7.77
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8
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9

cmkim/1로만들기_217p.py renamed to cmkim/이것이 취업을 위한 코딩 테스트다/1로만들기_217p.py

File renamed without changes.

cmkim/AdjacencyList1_2.py renamed to cmkim/이것이 취업을 위한 코딩 테스트다/AdjacencyList1_2.py

File renamed without changes.

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