zoukankan      html  css  js  c++  java
  • 【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
  • 相关阅读:
    gitlab
    MySQL千万级别大表,你要如何优化?
    kafka入门
    zookeeper的原理和应用
    MySQL 性能优化之慢查询
    Redis一些新的看法
    mysql 数据库锁
    MYSQL查看进程和kill进程
    hadoop批量命令脚本xcall.sh及jps找不到命令解决
    java stream 处理分组后取每组最大
  • 原文地址:https://www.cnblogs.com/seyjs/p/10636834.html
Copyright © 2011-2022 走看看