【leetcode】1020. Number of Enclaves

题目如下：

Given a 2D array A, 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 we cannot walk 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. 1 <= A.length <= 500
2. 1 <= A[i].length <= 500
3. 0 <= A[i][j] <= 1
4. All rows have the same size.

解题思路：本题和以前的岛屿问题是一样的，用DFS/BFS就好了。

代码如下：

class Solution(object):
    def numEnclaves(self, A):
        """
        :type A: List[List[int]]
        :rtype: int
        """
        direction = [(1,0),(-1,0),(0,1),(0,-1)]
        count_0 = 0
        count_2 = 0
        for i in range(len(A)):
            for j in range(len(A[i])):
                if A[i][j] == 0:
                    count_0 += 1
                    continue
                elif (i == 0 or j == 0 or i == len(A) - 1 or j == len(A[0])-1) and A[i][j] == 1:
                    A[i][j] = 2
                    queue = [(i,j)]
                    count_2 += 1
                    while len(queue) > 0:
                        a,b = queue.pop(0)
                        for (x,y) in direction:
                            if (x + a >= 0) and (x+ a < len(A)) and (y + b >= 0) and (y+b < len(A[i])) and A[x+a][y+b] == 1:
                                A[x + a][y + b] = 2
                                count_2 += 1
                                queue.append((x+a,y+b))
        #print A
        return len(A) * len(A[0]) - count_0 - count_2