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★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ Given a 2D array 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:
给出一个二维数组 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:
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 }
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