[Swift]LeetCode1020. 飞地的数量 | Number of Enclaves
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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 <= A.length <= 5001 <= A[i].length <= 5000 <= A[i][j] <= 1- All rows have the same size.
给出一个二维数组 A，每个单元格为 0（代表海）或 1（代表陆地）。
移动是指在陆地上从一个地方走到另一个地方（朝四个方向之一）或离开网格的边界。
返回网格中无法在任意次数的移动中离开网格边界的陆地单元格的数量。
示例 1：
输入：[[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]] 输出：3 解释： 有三个 1 被 0 包围。一个 1 没有被包围，因为它在边界上。
示例 2：
输入：[[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]] 输出：0 解释： 所有 1 都在边界上或可以到达边界。
提示：
1 <= A.length <= 5001 <= A[i].length <= 5000 <= A[i][j] <= 1- 所有行的大小都相同
432ms
1 class Solution {
2 func numEnclaves(_ A: [[Int]]) -> Int {
3 var grid = A
4 for row in 0..<grid.count {
5 dfs(&grid, row, 0)
6 dfs(&grid, row, grid[0].count - 1)
7 }
8
9 if grid.count > 0 && grid[0].count > 0 {
10 for col in 0..<grid[0].count {
11 dfs(&grid, 0, col)
12 dfs(&grid, grid.count - 1, col)
13 }
14 }
15
16 var result = 0
17 for row in 0..<grid.count {
18 for col in 0..<grid[0].count {
19 result += grid[row][col]
20 }
21 }
22 return result
23 }
24
25 func dfs(_ grid: inout [[Int]], _ row: Int, _ col: Int) {
26 if grid.count == 0 || grid[0].count == 0 {
27 return
28 }
29 if row < 0 || row > grid.count - 1
30 || col < 0 || col > grid[0].count - 1 {
31 return
32 }
33
34 if grid[row][col] == 0 {
35 return
36 }
37
38 grid[row][col] = 0
39
40 dfs(&grid, row + 1, col);
41 dfs(&grid, row - 1, col);
42 dfs(&grid, row, col + 1);
43 dfs(&grid, row, col - 1);
44 }
45 }444ms
1 class Solution {
2 func numEnclaves(_ A: [[Int]]) -> Int {
3 guard A.count > 0 else { return 0 }
4 guard A[0].count > 0 else { return 0 }
5
6 let h = A.count
7 let w = A[0].count
8
9 var visited = Array(repeating: Array(repeating: false, count: w), count: h)
10 var queue = [(Int, Int)]()
11 for i in 0..<h {
12 if A[i][0] == 1 {
13 queue.append((i, 0))
14 visited[i][0] = true
15 }
16
17 if w != 1 && A[i][w - 1] == 1 {
18 queue.append((i, w - 1))
19 visited[i][w - 1] = true
20 }
21 }
22
23 for j in 0..<w {
24 if A[0][j] == 1 {
25 queue.append((0, j))
26 visited[0][j] = true
27 }
28
29 if h != 1 && A[h - 1][j] == 1 {
30 queue.append((h - 1, j))
31 visited[h - 1][j] = true
32 }
33 }
34
35 while queue.count > 0 {
36 var nextQueue = [(Int, Int)]()
37 for point in queue {
38 let row = point.0
39 let col = point.1
40
41 if row - 1 >= 0 && A[row - 1][col] == 1 && !visited[row - 1][col] {
42 visited[row - 1][col] = true
43 nextQueue.append((row - 1, col))
44 }
45
46 if row + 1 < h && A[row + 1][col] == 1 && !visited[row + 1][col] {
47 visited[row + 1][col] = true
48 nextQueue.append((row + 1, col))
49 }
50
51 if col - 1 >= 0 && A[row][col - 1] == 1 && !visited[row][col - 1] {
52 visited[row][col - 1] = true
53 nextQueue.append((row, col - 1))
54 }
55
56 if col + 1 < w && A[row][col + 1] == 1 && !visited[row][col + 1] {
57 visited[row][col + 1] = true
58 nextQueue.append((row, col + 1))
59 }
60 }
61 queue = nextQueue
62 }
63
64 var count = 0
65
66 for i in 0..<h {
67 for j in 0..<w {
68 if A[i][j] == 1 && !visited[i][j] {
69 count += 1
70 }
71 }
72 }
73
74 return count
75 }
76 }452ms
1 class Solution {
2 var sum = 0
3 var ovSum = 0
4
5 func numEnclaves(_ A: [[Int]]) -> Int {
6 var a = A
7
8 var rowInd = 0
9 var colInd = 0
10 print(ovSum, sum)
11 rowInd = 0
12 while rowInd < A.count {
13 defer { rowInd += 1 }
14 colInd = 0
15 while colInd < A[rowInd].count {
16 defer { colInd += 1 }
17 if a[rowInd][colInd] == 1 {
18 ovSum += 1
19 }
20 }
21 }
22
23 for i in (0..<A.count) {
24 if a[i][0] == 1 { dfs(&a, i, 0) }
25 if a[i][A[0].count-1] == 1 { dfs(&a, i, A[0].count-1)}
26
27 }
28 for i in (0..<A[0].count) {
29 if a[0][i] == 1 { dfs(&a, 0, i) }
30 if a[A.count-1][i] == 1 { dfs(&a, A.count-1, i) }
31
32 }
33
34 print(ovSum, sum)
35 return ovSum - sum
36 }
37
38 func dfs(_ a: inout [[Int]], _ rowInd: Int, _ colInd: Int) {
39 guard rowInd < a.count, colInd < a[0].count, rowInd >= 0, colInd >= 0 else { return }
40 if a[rowInd][colInd] != 1 { return }
41 a[rowInd][colInd] = 2; sum += 1
42 dfs(&a, rowInd - 1, colInd)
43 dfs(&a, rowInd + 1, colInd)
44 dfs(&a, rowInd, colInd - 1)
45 dfs(&a, rowInd, colInd + 1)
46 }
47 }Runtime: 488 ms
Memory Usage: 19.1 MB
1 class Solution {
2 var DR:[Int] = [-1, 0, +1, 0]
3 var DC:[Int] = [0, +1, 0, -1]
4 var R:Int = 0
5 var C:Int = 0
6 var grid:[[Int]] = [[Int]]()
7 var visited:[[Bool]] = [[Bool]](repeating:[Bool](repeating:false,count:505),count:505)
8
9 func numEnclaves(_ A: [[Int]]) -> Int {
10 grid = A
11 R = grid.count
12 C = grid[0].count
13
14 for r in 0..<R
15 {
16 for c in 0..<C
17 {
18 if r == 0 || r == R - 1 || c == 0 || c == C - 1
19 {
20 if grid[r][c] == 1 && !visited[r][c]
21 {
22 dfs(r, c)
23 }
24 }
25 }
26 }
27 var ans:Int = 0
28 for r in 0..<R
29 {
30 for c in 0..<C
31 {
32 if grid[r][c] == 1 && !visited[r][c]
33 {
34 ans += 1
35 }
36 }
37 }
38 return ans
39 }
40
41 func dfs(_ r:Int,_ c:Int)
42 {
43 visited[r][c] = true
44 for dir in 0..<4
45 {
46 var nr:Int = r + DR[dir]
47 var nc:Int = c + DC[dir]
48 if nr >= 0 && nr < R && nc >= 0 && nc < C
49 {
50 if grid[nr][nc] == 1 && !visited[nr][nc]
51 {
52 dfs(nr, nc)
53 }
54 }
55 }
56 }
57 }524ms
1 class Solution
2 {
3 func numEnclaves(_ A: [[Int]]) -> Int
4 {
5 guard A.count > 0 else { return 0 }
6
7 var m = A
8 var ret = 0
9 for r in 0..<m.count
10 {
11 for c in 0..<m[r].count
12 {
13 var temp = 0
14 self.dfs(r,c, &m, &temp)
15 if temp != -1 { ret += temp}
16 }
17 }
18
19 return ret
20 }
21
22 // checked: -1
23 private func dfs(_ r: Int, _ c: Int, _ m: inout [[Int]], _ count: inout Int)
24 {
25 guard r >= 0, c >= 0, r < m.count, c < m[r].count, m[r][c] != -1 else { return }
26
27 if m[r][c] == 0 {
28 m[r][c] = -1
29 return
30 }
31
32 if r == 0 || c == 0 || r == m.count - 1 || c == m[r].count - 1 { count = -1 }
33 if count != -1 { count += 1 }
34 m[r][c] = -1
35 // up
36 if r > 0 { self.dfs(r - 1, c, &m, &count) }
37 // down
38 if r < m.count - 1 { self.dfs(r + 1, c, &m, &count) }
39 // left
40 if c > 0 { self.dfs(r, c - 1, &m, &count) }
41 // right
42 if c < m[r].count - 1 { self.dfs(r, c + 1, &m, &count) }
43 }
44 }