Number of Enclaves

Given a 2D arrayA, each cell is 0 (representing sea) or 1 (representing land)

A move consists of walking from one land square 4-directionally to another land square, or off the boundary of the grid.

Return the number of land squares in the grid for which wecannotwalk off the boundary of the grid in any number of moves.

Example 1:

Input: 
[[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
Output: 
3
Explanation: 

There are three 1s that are enclosed by 0s, and one 1 that isn't enclosed because its on the boundary.

Example 2:

Input: 
[[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
Output: 
0
Explanation: 

All 1s are either on the boundary or can reach the boundary.

Note:

1 <= A.length <= 500
1 <= A[i].length <= 500
0 <= A[i][j] <= 1
All rows have the same size.

分析

从边缘1开始往内不断扩张并flip，最后数剩下1就是答案

class Solution:
    def numEnclaves(self, A: List[List[int]]) -> int:
        if not A or not A[0]:
            return 0
        res = float('inf')
        d = [-1, 0, 1, 0, -1]
        n, m = len(A), len(A[0])
        def dfs(x, y):           
            if A[x][y] == -1:
                return            
            A[x][y] = -1
            for nx, ny in [(x + d[k], y + d[k + 1]) for k in range(4)]:
                if 0 <= nx < n and 0 <= ny < m and A[nx][ny] == 1:
                    dfs(nx, ny)

        for i in range(n):
            for j in range(m):
                if (i==0 or i == n-1 or j == 0 or j == m-1) and A[i][j] == 1:
                    dfs(i,j)
        cnt = 0            
        for i in range(n):
            for j in range(m):
                if A[i][j] == 1:
                    cnt += 1
        return cnt

BFS

class Solution:
    def numEnclaves(self, A: List[List[int]]) -> int:
        if not A or not A[0]:
            return 0
        res = float('inf')
        d = [-1, 0, 1, 0, -1]
        n, m = len(A), len(A[0])

        for i in range(n):
            for j in range(m):
                if (i==0 or i == n-1 or j == 0 or j == m-1) and A[i][j] == 1:
                    q = [(i,j)]
                    while q:
                        x,y=q.pop()
                        if A[x][y] == -1:
                            continue            
                        A[x][y] = -1
                        for nx, ny in [(x + d[k], y + d[k + 1]) for k in range(4)]:
                            if 0 <= nx < n and 0 <= ny < m and A[nx][ny] == 1:
                                q.append((nx, ny))
        cnt = 0            
        for i in range(n):
            for j in range(m):
                if A[i][j] == 1:
                    cnt += 1
        return cnt

BFS基础上再加union find

class Solution:
    def numEnclaves(self, A: List[List[int]]) -> int:
        if not A or not A[0]:
            return 0
        res = float('inf')
        d = [-1, 0, 1, 0, -1]
        n, m = len(A), len(A[0])

        f={}
        def find(x):
            f.setdefault(x,x)
            if f[x]!=x:
                f[x] = find(f[x])
            return f[x]
        def union(x,y):
            f[find(x)]=find(y)

        for i in range(n):
            for j in range(m):
                if (i==0 or i == n-1 or j == 0 or j == m-1) and A[i][j] == 1:
                    q= [(i,j)]
                    while q:
                        x,y=q.pop()
                        if A[x][y] == -1:
                            continue
                        A[x][y] = -1
                        for nx, ny in [(x + d[k], y + d[k + 1]) for k in range(4)]:
                            if 0 <= nx < n and 0 <= ny < m and A[nx][ny] == 1:
                                q.append((nx, ny))
                                union((nx, ny),(i,j))

        cnt = 0            
        for i in range(n):
            for j in range(m):
                if A[i][j] == 1 and (i,j) not in f:
                    cnt += 1
        return cnt