DAY27-T1984&T56-2022-02-11-非自己作答

class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        vis = [[False] * n for _ in range(m)]

        def dfs(r: int, c: int) -> None:
            if r < 0 or r >= m or c < 0 or c >= n or grid[r][c] == 0 or vis[r][c]:
                return
            vis[r][c] = True
            for x, y in ((r - 1, c), (r + 1, c), (r, c - 1), (r, c + 1)):
                dfs(x, y)

        for i in range(m):
            dfs(i, 0)
            dfs(i, n - 1)
        for j in range(1, n - 1):
            dfs(0, j)
            dfs(m - 1, j)
        return sum(grid[i][j] and not vis[i][j] for i in range(1, m - 1) for j in range(1, n - 1))

# 作者：LeetCode-Solution
# 链接：https://leetcode-cn.com/problems/number-of-enclaves/solution/fei-di-de-shu-liang-by-leetcode-solution-nzs3/

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        intervals.sort(key=lambda x: x[0])

        merged = []
        for interval in intervals:
            # 如果列表为空，或者当前区间与上一区间不重合，直接添加
            if not merged or merged[-1][1] < interval[0]:
                merged.append(interval)
            else:
                # 否则的话，我们就可以与上一区间进行合并
                merged[-1][1] = max(merged[-1][1], interval[1])

        return merged

# 作者：LeetCode-Solution
# 链接：https://leetcode-cn.com/problems/merge-intervals/solution/he-bing-qu-jian-by-leetcode-solution/