Q22.블록이동하기 다시풀기

코딩테스트 연습 - 블록 이동하기 | 프로그래머스 (programmers.co.kr)

코딩테스트 연습 - 블록 이동하기

 

손쉽게 풀었다. 코딩 실력이 늘고 있는게 느껴지는구만

이 문제는 좌표에 유의해서 풀어야 한다.

 

from collections import deque

def solution(board):
    N = len(board)
    newboard = [[1] * (N+2) for _ in range(N+2)]
    for i in range(N):
        for j in range(N):
            newboard[i+1][j+1] = board[i][j]
            
    startpos = {(1,1),(1,2)}
    
    q = deque()
    q.append((startpos,0))
    visited = []
    
    answer = 0
    while q:
        position, distance = q.popleft()
        
        if position in visited:
            continue
        else:
            visited.append(position)
            
        print(position)
        
        position = list(position)
        if (N,N) in position:
            answer = distance
            break
        
        x1,y1 = position[0]
        x2,y2 = position[1]
        
        if y1==y2:
            if newboard[x1][y1+1]==0 and newboard[x2][y2+1]==0:
                q.append(({(x1,y1),(x1,y1+1)}, distance+1))
                q.append(({(x2,y2),(x2,y2+1)}, distance+1))
            if newboard[x1][y1-1]==0 and newboard[x2][y2-1]==0:
                q.append(({(x1,y1),(x1,y1-1)}, distance+1))
                q.append(({(x2,y2),(x2,y2-1)}, distance+1))
        elif x1==x2:
            if newboard[x1+1][y1]==0 and newboard[x2+1][y2]==0:
                q.append(({(x1,y1),(x1+1,y1)}, distance+1))
                q.append(({(x2,y2),(x2+1,y2)}, distance+1))
            if newboard[x1-1][y1]==0 and newboard[x2-1][y2]==0:
                q.append(({(x1,y1),(x1-1,y1)}, distance+1))
                q.append(({(x2,y2),(x2-1,y2)}, distance+1))
        dx = [0,0,-1,1]
        dy = [1,-1,0,0]
        for i in range(4):
            newx1 = x1+dx[i]
            newy1 = y1+dy[i]
            newx2 = x2+dx[i]
            newy2 = y2+dy[i]
            
            if newboard[newx1][newy1]==0 and newboard[newx2][newy2]==0:
                q.append( ({(newx1,newy1),(newx2,newy2)}, distance+1) )
            
    return answer