[Swift]LeetCode200.岛屿的个数 | Number of Islands

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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

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:

示例 2:

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 }

posted @ 2018-12-26 18:39 为敢技术阅读(583) 评论(0) 编辑收藏举报

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[Swift]LeetCode200.岛屿的个数 | Number of Islands

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