Merge branch 'master' of https://github.com/CharryWu/leetcode

Author: CharryWu

simulation/289. Game of Life.py

@@ -0,0 +1,32 @@
+class Solution:
+    def gameOfLife(self, board: List[List[int]]) -> None:
+        """
+        Use 4 values, instead of 2 (0/1), to designate the state after update. This allows us to retain the prev state information so updates of current cell won't affect neighbor cell calculations
+        1: live -> live cell (no change)
+        -1: live -> dead cell
+        check abs(nei) == 1: whether nei cell is previously alive (so update doesn't affect calculating its neighbors' next state)
+        0: dead -> dead cell (no change)
+        2: dead -> live cell (doens't affect the abs(nei) check)
+
+        After ALL cells have been updated, check if cell is > 0. if > 0, the values are 1, 2, then the post-update cell is alive, otherwise the values are 0, -1, dead
+        """
+        directions = [(1,0), (1,-1), (0,-1), (-1,-1), (-1,0), (-1,1), (0,1), (1,1)]
+        m, n = len(board), len(board[0])
+
+        for x in range(m):
+            for y in range(n):
+                live = 0
+                for dx, dy in directions: # count live neighbors
+                    nx, ny = x+dx, y+dy
+                    if 0 <= nx < m and 0 <= ny < n and abs(board[nx][ny]) == 1:
+                        live += 1
+                if board[x][y] == 1 and (live < 2 or live > 3): # Rule 1 or Rule 3: live -> dead
+                    board[x][y] = -1
+                if board[x][y] == 0 and live == 3: # Rule 4: dead -> live
+                    board[x][y] = 2
+                # else: Rule 2 no change: dead -> dead, live -> live
+
+        for x in range(m):
+            for y in range(n):
+                board[x][y] = 1 if board[x][y] > 0 else 0
+        return board