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  • [Swift]LeetCode1254. 统计封闭岛屿的数目 | Number of Closed Islands

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    Given a 2D grid consists of 0s (land) and 1s (water).  An island is a maximal 4-directionally connected group of 0s and a closed island is an island totally (all left, top, right, bottom) surrounded by 1s.

    Return the number of closed islands

    Example 1:

    Input: grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
    Output: 2
    Explanation: 
    Islands in gray are closed because they are completely surrounded by water (group of 1s).

    Example 2:

    Input: grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
    Output: 1
    

    Example 3:

    Input: grid = [[1,1,1,1,1,1,1],
                   [1,0,0,0,0,0,1],
                   [1,0,1,1,1,0,1],
                   [1,0,1,0,1,0,1],
                   [1,0,1,1,1,0,1],
                   [1,0,0,0,0,0,1],
                   [1,1,1,1,1,1,1]]
    Output: 2

    Constraints:

    • 1 <= grid.length, grid[0].length <= 100
    • 0 <= grid[i][j] <=1

    有一个二维矩阵 grid ,每个位置要么是陆地(记号为 0 )要么是水域(记号为 1 )。

    我们从一块陆地出发,每次可以往上下左右 4 个方向相邻区域走,能走到的所有陆地区域,我们将其称为一座「岛屿」。

    如果一座岛屿 完全 由水域包围,即陆地边缘上下左右所有相邻区域都是水域,那么我们将其称为 「封闭岛屿」。

    请返回封闭岛屿的数目。

    示例 1:

    输入:grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
    输出:2
    解释:
    灰色区域的岛屿是封闭岛屿,因为这座岛屿完全被水域包围(即被 1 区域包围)。

    示例 2:

    输入:grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
    输出:1
    

    示例 3:

    输入:grid = [[1,1,1,1,1,1,1],
                 [1,0,0,0,0,0,1],
                 [1,0,1,1,1,0,1],
                 [1,0,1,0,1,0,1],
                 [1,0,1,1,1,0,1],
                 [1,0,0,0,0,0,1],
                 [1,1,1,1,1,1,1]]
    输出:2

    提示:

    • 1 <= grid.length, grid[0].length <= 100
    • 0 <= grid[i][j] <=1

    Runtime: 92 ms
    Memory Usage: 21.1 MB
     1 class Solution {
     2     func closedIsland(_ grid: [[Int]]) -> Int {
     3         var grid = grid
     4         for i in 0..<grid.count
     5         {
     6             for j in 0..<grid[i].count
     7             {
     8                 if i == 0 || j == 0 || i == grid.count - 1 || j == grid[i].count - 1
     9                 {
    10                     fill(&grid, i, j)
    11                 }
    12             }
    13         }
    14         var res:Int = 0
    15         for i in 0..<grid.count
    16         {
    17             for j in 0..<grid[i].count
    18             {
    19                 res += fill(&grid, i, j) > 0 ? 1 : 0
    20             }
    21         }
    22         return res
    23     }
    24     
    25     func fill(_ g: inout [[Int]],_ i:Int,_ j:Int) -> Int
    26     {
    27         if i < 0 || j < 0 || i >= g.count || j >= g[i].count || g[i][j] != 0
    28         {
    29             return 0
    30         }
    31         g[i][j] = 1
    32         return g[i][j] + fill(&g, i + 1, j) + fill(&g, i, j + 1) + fill(&g, i - 1, j) + fill(&g, i, j - 1)
    33     }
    34 }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/11831493.html
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