목록289. Game of Life (1)

Scribbling

LeetCode: 289. Game of Life

To solve this problem in-place, we can think of using two more states(2, 3) rather than two states(1, 0). class Solution: def gameOfLife(self, board: List[List[int]]) -> None: """ Do not return anything, modify board in-place instead. """ m, n = len(board), len(board[0]) for y in range(m): for x in range(n): cnt = self.countOnes(board, y, x) if board[y][x] == 0: if cnt == 3: board[y][x] = 3 else..
Computer Science/Coding Test 2021. 12. 29. 20:18
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