feat:구현

Author: Suxxxxhyun

Week5/공통/키패드누르기/suhyun.py

@@ -0,0 +1,52 @@
+# 3:51 ~ 
+# Hint : 1부터 0까지 좌표를 저장하는 딕셔너리를 선언한다.
+
+pad = {
+    '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),
+    '0' : (3,1),
+    '*' : (3,0),
+    '#' : (3,2)
+}
+
+def solution(numbers, hand):
+    answer = ''
+    
+    left = pad['*']
+    right = pad['#']
+    
+    for n in numbers:
+        if n in [1, 4, 7]:
+            answer += 'L'
+            left = pad[str(n)]
+        if n in [3, 6, 9]:
+            answer += 'R'
+            right = pad[str(n)]
+        if n in [2,5,8,0]:
+            L1 = abs(pad[str(n)][0] - left[0]) + abs(pad[str(n)][1] - left[1])
+            R1 = abs(pad[str(n)][0] - right[0]) + abs(pad[str(n)][1] - right[1])
+            if L1 == R1:
+                if hand == "right":
+                    right = pad[str(n)]
+                    answer += "R"
+                else:
+                    left = pad[str(n)]
+                    answer += "L"
+            elif L1 > R1:
+                answer += "R"
+                right = pad[str(n)]
+            else:
+                answer += "L"
+                left = pad[str(n)]
+    return answer
+
+numbers = [0,0]
+hand = "right"
+print(solution(numbers, hand))