Re: [閒聊] 每日leetcode

[dont (dont)]

※ 引述《JIWP (神楽めあ的錢包)》之銘言：
: 2290. Minimum Obstacle Removal to Reach Corner
: 思路:
: (2) BFS

可以直接用一個Deque (cost, r, c)
grid[i][j]=0就加到前面
grid[i][j]=1就加到後面

```python
class Solution:
    def minimumObstacles(self, grid: List[List[int]]) -> int:
        len_r, len_c = len(grid), len(grid[0])

        grid[0][0] = -1
        queue = deque([(0, 0, 0)])  # cost, r, c

        while queue:
            cost, r, c = queue.popleft()
            if r == len_r-1 and c == len_c-1:
                return cost
            for nr, nc in (r+1, c), (r-1, c), (r, c+1), (r, c-1):
                if 0 <= nr < len_r and 0 <= nc < len_c and grid[nr][nc] != -1:
                    if grid[nr][nc] == 0:
                        queue.appendleft((cost, nr, nc))
                    else:
                        queue.append((cost+1, nr, nc))
                    grid[nr][nc] = -1
```

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推 cities516: Deque 超方便的