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太平洋大西洋水流问题

这道题只能写逆流！！顺流试了好几次内存都不够，所以只能从海洋出发，然后看那些地方是可以达到的，这是见过的第二个逆流问题了，上一题是130题。
写法上这个dfs没有return，所以就是其实回溯不需要明确的终止，如果需要的话走完也可以。

class Solution:
    def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:
        def dfs(i,j,matrix):
            matrix[i][j] = True 
            for k1,k2 in [[i-1,j],[i+1,j],[i,j-1],[i,j+1]]:
                if (0 <= k1 < m and 0 <= k2 < n) and matrix[k1][k2] == True: continue
                if (0 <= k1 < m and 0 <= k2 < n) and heights[k1][k2] >=  heights[i][j]:
                    dfs(k1,k2,matrix)
        
        m = len(heights)
        n = len(heights[0])
        Pa = [[False]*n for _ in range(m)]
        At = [[False]*n for _ in range(m)]
        res = []
        for i in range(m):
            dfs(i,0,Pa)
            dfs(i,n-1,At)
        for j in range(n):
            dfs(0,j,Pa)
            dfs(m-1,j,At)
        
        for i in range(m):
            for j in range(n):
                if Pa[i][j] and At[i][j]:
                    res.append([i,j])
        return res

飞地的数量

和130是一样的题，也是从外到里比较好做。

class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        def dfs(i,j):
            grid[i][j] = 0
            for k1,k2 in [[i-1,j],[i+1,j],[i,j+1],[i,j-1]]:
                if (0 <= k1 < m and 0<= k2 < n) and grid[k1][k2] == 1:
                    dfs(k1,k2)
        m = len(grid)
        n = len(grid[0])

        for i in range(m):
            if grid[i][0] == 1: 
                dfs(i,0)
            if grid[i][n-1] == 1: 
                dfs(i,n-1)
        
        for j in range(n):
            if grid[0][j] == 1: 
                dfs(0,j)
            if grid[m-1][j] == 1: 
                dfs(m-1,j)
        
        return sum(sum(grid[i]) for i in range(m))

统计封闭岛屿的数目

一样的问题顺手做了，我这里的思路就是先把没有被包围的岛（0）转成水（1），然后再去找被包围的岛。

class Solution:
    def closedIsland(self, grid: List[List[int]]) -> int:
        def dfs(i,j):
            grid[i][j] = 1
            for k1,k2 in [[i-1,j],[i+1,j],[i,j+1],[i,j-1]]:
                if (0 <= k1 < m and 0<= k2 < n) and grid[k1][k2] == 0:
                    dfs(k1,k2)
        m = len(grid)
        n = len(grid[0])
        for i in range(m):
            if grid[i][0] == 0: 
                dfs(i,0)
            if grid[i][n-1] == 0: 
                dfs(i,n-1)
        
        for j in range(n):
            if grid[0][j] == 0: 
                dfs(0,j)
            if grid[m-1][j] == 0: 
                dfs(m-1,j)
        #print(grid)

        res = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 0:
                    dfs(i,j)
                    res += 1
        return res