programmers- lv.1 (키패드 누르기)

문제링크: 키패드 누르기

✍🏻 Information

	content
언어	python
난이도	⭐️⭐️
풀이시간	30분
제출횟수	2
인터넷검색유무	yes

🍒 My Code

phone = [[1,2,3],
         [4,5,6],
         [7,8,9],
         ['*',0,'#']]
def check_distance(x,y,target_x,target_y):
    return abs(x-target_x)+abs(y-target_y)
def solution(numbers, hand):
    answer = ''
    lx,ly = 3, 0
    rx,ry = 3, 2
    for number in numbers:
        coordinate=[[i,j] for i in range(4) for j in range(3) if phone[i][j]==number] #핸드폰에서 현재 좌표찾기
        print(coordinate)
        
        if coordinate[0][1]==0:
            answer+='L'
            lx,ly = coordinate[0][0],coordinate[0][1]
        
        elif coordinate[0][1]==1:
            left = check_distance(lx,ly,coordinate[0][0],coordinate[0][1])
            right = check_distance(rx,ry,coordinate[0][0],coordinate[0][1])
            if left<right:
                answer+='L'
                lx,ly = coordinate[0][0],coordinate[0][1]
            elif left>right:
                answer+='R'
                rx,ry = coordinate[0][0],coordinate[0][1]
            else:
                if hand=="left":
                    answer+='L'
                    lx,ly = coordinate[0][0],coordinate[0][1]
                else:
                    answer+='R'
                    rx,ry = coordinate[0][0],coordinate[0][1]
        elif coordinate[0][1]==2:
            answer+='R'
            rx,ry = coordinate[0][0],coordinate[0][1]
        
    return answer

💡 What I learned

• 2차원 list에서 원하는 value의 index 찾기
  : [ (i, j) ]형태로 담긴다.

newlist=[(i,j) for i in range(n) for j in range(m) if mylist[i][j]==1]

이렇게 하니 coordinate[0]을 다 붙여줘야하는 귀찮음이 있었는데 해결할 방법을 모색해봐야겠다.

• 다른 사람 풀이

<풀이1 - 좌표를 지정>

def solution(numbers, hand):
    answer = ''
    location = [[3, 1], [0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2], [2, 0], [2, 1], [2, 2]]
    left, right = [3, 0], [3, 2]
    for i in numbers:
        if i % 3 == 1:
            answer += 'L'
            left = location[i]
        elif i % 3 == 0 and i != 0:
            answer += 'R'
            right = location[i]
        else:
            l = abs(location[i][0] - left[0]) + abs(location[i][1] - left[1])
            r = abs(location[i][0] - right[0]) + abs(location[i][1] - right[1])
            if l < r:
                answer += 'L'
                left = location[i]
            elif l > r:
                answer += 'R'
                right = location[i]
            else:
                answer += hand[0].upper() #L이나 R을 answer에 더하는 과정
                if hand == 'right':
                    right = location[i]
                else:
                    left = location[i]                

    return answer

<풀이2 - 딕셔너리 사용>

def solution(numbers, hand):
    answer = ''
    key_dict = {1:(0,0),2:(0,1),3:(0,2),
                4:(1,0),5:(1,1),6:(1,2),
                7:(2,0),8:(2,1),9:(2,2),
                '*':(3,0),0:(3,1),'#':(3,2)}

    left = [1,4,7]
    right = [3,6,9]
    lhand = '*'
    rhand = '#'
    for i in numbers:
        if i in left:
            answer += 'L'
            lhand = i
        elif i in right:
            answer += 'R'
            rhand = i
        else:
            curPos = key_dict[i]
            lPos = key_dict[lhand]
            rPos = key_dict[rhand]
            ldist = abs(curPos[0]-lPos[0]) + abs(curPos[1]-lPos[1])
            rdist = abs(curPos[0]-rPos[0]) + abs(curPos[1]-rPos[1])

            if ldist < rdist:
                answer += 'L'
                lhand = i
            elif ldist > rdist:
                answer += 'R'
                rhand = i
            else:
                if hand == 'left':
                    answer += 'L'
                    lhand = i
                else:
                    answer += 'R'
                    rhand = i

    return answer