1254. Number of Closed Islands ⭐

Medium
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

解題

經典 DFS 題目
Runtime: 12 ms, faster than 87.76%
Memory Usage: 4.6 MB, less than 85.71%
func closedIsland(grid [][]int) int {
res := 0
​
for i := 0; i < len(grid); i++ {
for j := 0; j < len(grid[0]); j++ {
if grid[i][j] == 0 {
if dfs(grid, i, j) { res++ }
}
}
}
​
return res
}
​
func dfs(grid [][]int, x int, y int) bool {
if x < 0 || y < 0 || x == len(grid) || y == len(grid[0]) {
return false
}
​
if grid[x][y] == 1 { return true }
grid[x][y] = 1
​
right := dfs(grid, x + 1, y)
left := dfs(grid, x - 1, y)
up := dfs(grid, x, y + 1)
down := dfs(grid, x, y - 1)
​
return right && left && up && down
}
​