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➤微信公众号：山青咏芝（shanqingyongzhi）
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Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
Example 1:
Input: 11110 11010 11000 00000 Output: 1
Example 2:
Input: 11000 11000 00100 00011 Output: 3
Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
Example 1:
Input: 11110 11010 11000 00000 Output: 1
Example 2:
Input: 11000 11000 00100 00011 Output: 3
给定一个由 '1'（陆地）和 '0'（水）组成的的二维网格，计算岛屿的数量。一个岛被水包围，并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。
示例 1:
输入: 11110 11010 11000 00000 输出: 1
示例 2:
输入: 11000 11000 00100 00011 输出: 3
260ms
1 class Solution { 2 func numIslands(_ grid: [[Character]]) -> Int { 3 var map = grid 4 let row = map.count 5 if row == 0 {return 0} 6 let col = map[0].count 7 var count = 0 8 for i in 0..<row { 9 for j in 0..<col { 10 if map[i][j] == "1" { 11 dfs(&map, i, j, row, col) 12 count += 1 13 } 14 } 15 } 16 return count 17 } 18 func dfs(_ map: inout [[Character]], _ i: Int, _ j: Int, _ row: Int, _ col: Int) { 19 if (i >= 0 && i < row && j >= 0 && j < col && map[i][j] == "1") { 20 map[i][j] = "2" 21 dfs(&map, i, j-1, row, col); 22 dfs(&map, i-1, j, row, col); 23 dfs(&map, i, j+1, row, col); 24 dfs(&map, i+1, j, row, col); 25 } 26 } 27 }
268ms
1 class Solution { 2 func numIslands(_ grid: [[Character]]) -> Int { 3 var map = grid 4 let row = map.count 5 if row == 0 {return 0} 6 let col = map[0].count 7 var count = 0 8 for i in 0..<row { 9 for j in 0..<col { 10 if map[i][j] == "1" { 11 dfs(&map, i, j, row, col) 12 count += 1 13 } 14 } 15 } 16 return count 17 } 18 func dfs(_ map: inout [[Character]], _ i: Int, _ j: Int, _ row: Int, _ col: Int) { 19 if (i >= 0 && i < row && j >= 0 && j < col && map[i][j] == "1") { 20 map[i][j] = "2" 21 let index = [0,-1,0,1,0] 22 for k in 0..<(index.count - 1) { 23 dfs(&map, i + index[k], j + index[k + 1], row, col) 24 } 25 } 26 } 27 }
288ms
1 class Solution { 2 func numIslands(_ grid: [[Character]]) -> Int { 3 var grid = grid 4 var count = 0 5 6 for row in grid.indices { 7 for col in 0..<grid[row].count where grid[row][col] == "1" { //find the first point of contact 8 9 //dfs(grid: &grid, row: row, col: col) 10 11 bfs(grid: &grid, row: row, col: col) 12 13 count = count + 1 // found an island 14 } 15 } 16 17 return count 18 } 19 20 func bfs(grid: inout [[Character]], row: Int, col: Int) { 21 // bfs 只call了一次，所以可以直接默认进来的是1 22 var queue = [(Int, Int)]() 23 queue.append((row, col)) 24 grid[row][col] = "0" 25 26 while !queue.isEmpty { 27 let pair = queue.removeFirst() //don't use dropFirst 28 29 let x = pair.0 30 let y = pair.1 31 32 //up part 33 if x > 0, grid[x - 1][y] == "1" { 34 grid[x - 1][y] = "0" // mark as visited 35 queue.append((x - 1, y)) 36 } 37 38 //down part 39 if x < grid.count - 1, grid[x + 1][y] == "1" { 40 grid[x + 1][y] = "0" // mark as visited 41 queue.append((x + 1, y)) 42 } 43 44 //left part 45 if y > 0, grid[x][y - 1] == "1" { 46 grid[x ][y - 1] = "0" // mark as visited 47 queue.append((x, y - 1)) 48 } 49 50 //right part 51 if y < grid[x].count - 1, grid[x][y + 1] == "1" { 52 grid[x ][y + 1] = "0" // mark as visited 53 queue.append((x, y + 1)) 54 } 55 } 56 } 57 58 func dfs(grid: inout [[Character]], row: Int, col: Int) { 59 //dfs: pre order tree: me left righ => me, up, down, left right 60 61 // me step 62 guard grid[row][col] == "1" else { return } 63 64 grid[row][col] = "0"// mark as visited, smart 65 66 //up part 67 if row > 0, grid[row - 1][col] == "1" { 68 dfs(grid: &grid, row: row - 1, col: col) 69 } 70 71 //down part 72 if row < grid.count - 1, grid[row + 1][col] == "1" { 73 dfs(grid: &grid, row: row + 1, col: col) 74 } 75 76 //left part 77 if col > 0, grid[row][col - 1] == "1" { 78 dfs(grid: &grid, row: row, col: col - 1) 79 } 80 81 //right part 82 if col < grid[row].count - 1, grid[row][col + 1] == "1" { 83 dfs(grid: &grid, row: row, col: col + 1) 84 } 85 86 } 87 }
292ms
1 class Solution { 2 func numIslands(_ grid: [[Character]]) -> Int { 3 guard !grid.isEmpty, !grid[0].isEmpty else { return 0 } 4 5 var queue = [[Int]]() 6 var grid = grid 7 var count = 0 8 9 for i in 0..<grid.count { 10 for j in 0..<grid[0].count { 11 if grid[i][j] == "1" { 12 queue.append([i, j]) 13 count += 1 14 markIsland(&grid, &queue) 15 } 16 } 17 } 18 return count 19 } 20 21 22 private let dc = [-1, 1, 0, 0] 23 private let dr = [0, 0, 1, -1] 24 25 private func markIsland(_ grid: inout [[Character]], _ queue: inout [[Int]]) { 26 while !queue.isEmpty { 27 let curr = queue.removeFirst() 28 for i in 0..<dc.count { 29 let row = curr[0] + dr[i] 30 let column = curr[1] + dc[i] 31 if row < 0 || row >= grid.count || column < 0 || column >= grid[0].count || grid[row][column] == "0" { 32 continue 33 } 34 grid[row][column] = "0" 35 queue.append([row, column]) 36 } 37 } 38 } 39 }
296ms
1 class Solution { 2 func numIslands(_ grid: [[Character]]) -> Int { 3 guard grid.count > 0, grid[0].count > 0 else { 4 return 0 5 } 6 var result = 0 7 8 var dirs = [[0, 1], [1, 0], [0, -1], [-1, 0]] 9 10 var memo = [[Bool]](repeating:[Bool](repeating:false, count:grid[0].count), count:grid.count) 11 12 func dfs(i:Int, j:Int) { 13 if i < 0 || j < 0 || i >= grid.count || j >= grid[0].count || memo[i][j] == true || grid[i][j] == "0" { 14 return 15 } 16 17 memo[i][j] = true 18 19 for dir in dirs { 20 dfs(i:i + dir[0], j: j + dir[1]) 21 } 22 } 23 24 for i in 0..<grid.count { 25 for j in 0..<grid[i].count { 26 if memo[i][j] == true || grid[i][j] == "0" { 27 continue 28 } else { 29 result += 1 30 dfs(i:i, j:j) 31 } 32 } 33 } 34 35 return result 36 } 37 }
452ms
1 class Solution { 2 3 func numIslands(_ grid: [[Character]]) -> Int { 4 5 var map = grid 6 let n = map.count; 7 if n == 0 { return 0 } 8 let m = map[0].count 9 var count = 0 10 11 func emptMap(_ map: inout [[Character]], i: Int, j: Int) { 12 if (i < 0 || j<0 || i>=n || j>=m || map[i][j] != "1") { return } 13 map[i][j] = "0"; 14 // 递归清除所有地点 15 emptMap(&map, i: i+1, j: j) 16 emptMap(&map, i: i-1, j: j) 17 emptMap(&map, i: i, j: j+1) 18 emptMap(&map, i: i, j: j-1) 19 } 20 21 for (i,v) in map.enumerated() { 22 for (j,_) in v.enumerated () { 23 if (map[i][j] == "1") { 24 emptMap(&map, i: i, j: j) 25 count = count + 1 26 } 27 } 28 } 29 30 return count 31 } 32 }