Number of Islands

#DFS #UnionFind

Question

Given a boolean 2D matrix, find the number of islands.

Notice

0 is represented as the sea, 1 is represented as the island. If two 1 is adjacent, we consider them in the same island. We only consider up/down/left/right adjacent.

Example Given graph:

[
  [1, 1, 0, 0, 0],
  [0, 1, 0, 0, 1],
  [0, 0, 0, 1, 1],
  [0, 0, 0, 0, 0],
  [0, 0, 0, 0, 1]
]

return 3.

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Related Problems

Medium Surrounded Regions Hard Number of Islands II

Analysis

思路一： 此题可以考虑用Union Find，不过更简单的是用BFS或者DFS。 其中DFS结合mark的方法最巧妙简单，n^2循环，扫描grid[i][j], 如果是island的，即grid[i][j] == true，则计数加一（ans++），并对四个方向进行DFS查找，并将所有属于那坐岛屿的点mark为非岛屿。这样扫描过的地方全部会变成非岛屿，而岛屿的数量已被记录。

思路二： Union Find 利用Union Find的find和union操作，对岛屿进行合并计数。因为UnionFind结构一般来说是一维的

| 1 | 2 | 3 | 4 |
-----------------
| 2 | 2 | 2 | 4 |

表达了如下的连通结构

1 - 2    4
| /
3

因此可以转换二维矩阵坐标为一维数字，M N 的矩阵中`(i,j) -> i N + j`。

1. 建立UnionFind的parent[] (或者 parent hashmap)，初始化各个parent[i * N + j] = i * N + j，如果grid[i][j] == true，则岛屿计数count++

2. 之后再对矩阵遍历一次，当遇到岛屿时，则要检查上下左右四个方向的邻近点，如果也是岛屿则合并，并且count--；

Union Find结构可以用Path Compression和Weighted Union进行优化。

Solution

DFS (3ms AC)

public class Solution {
    /**
     * @param grid a boolean 2D matrix
     * @return an integer
     */
    public int numIslands(boolean[][] grid) {
        m = grid.length;
        if (m == 0) {
            return 0;
        }
        n = grid[0].length;
        if (n == 0) {
            return 0;
        }
        int nums = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == false) {
                    continue;
                }
                nums++;
                dfs(grid, i, j);
            }
        }
        return nums;
    }

    private void dfs(boolean[][] grid, int i, int j) {
        if (i < 0 || i >= m || j < 0 || j >= n) {
            return;
        }
        if (grid[i][j] == true) {
            grid[i][j] = false;
            dfs(grid, i - 1, j);
            dfs(grid, i + 1, j);
            dfs(grid, i, j - 1);
            dfs(grid, i, j + 1);
        }
    }

    private int m, n;
}

DFS (same, just with directional 2D array)

class Solution {
    int[][] dirs = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    public int numIslands(char[][] grid) {
        int m = grid.length;
        if (m == 0) {
            return 0;
        }
        int n = grid[0].length;
        int result = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    dfs(grid, i, j, m, n);
                    result++;
                }
            }
        }
        return result;
    }
    public void dfs(char[][] grid, int i, int j, int m, int n) {
        if (i < 0 || i > m - 1 || j < 0 || j > n - 1 || grid[i][j] != '1') {
            return;
        }
        grid[i][j] = '0';
        for (int[] dir : dirs) {
            dfs(grid, i + dir[0], j + dir[1], m, n);
        }
    }
}

BFS - (14ms AC)

class Solution {
    int[][] dir = new int[][] { {1, 0}, {0, 1}, {-1, 0}, {0, -1} };

    class Point {
        int row;
        int col;
        Point (int row, int col) {
            this.row = row;
            this.col = col;
        }
    }

    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        int m = grid.length;
        int n = grid[0].length;
        int numIslands = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    numIslands++;
                    grid[i][j] = '0';
                    // BFS
                    Queue<Point> q = new LinkedList<>();
                    q.offer(new Point(i, j));
                    while (!q.isEmpty()) {
                        Point p = q.poll();
                        for (int k = 0; k < 4; k++) {
                            int ni = p.row + dir[k][0];
                            int nj = p.col + dir[k][1];
                            if (ni >= 0 && ni < m && 
                                nj >= 0 && nj < n && 
                                grid[ni][nj] == '1') {
                                grid[ni][nj] = '0';
                                q.offer(new Point(ni, nj));
                            }
                        }
                    }
                }
            }
        }
        return numIslands;
    }


}

BFS - (17ms AC)

class Solution {
    private int m, n;
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        m = grid.length;
        n = grid[0].length;
        int count = 0;
        boolean [][] visited = new boolean[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1' && !visited[i][j]) {
                    count++;
                    bfs(grid, visited, i, j);
                }
            }
        }
        return count; 
    }

    // BFS - marking island as visited
    private void bfs(char[][] grid, boolean[][] visited, int x, int y) {
        int[] dx = {1, 0 , -1, 0};
        int[] dy = {0, 1 , 0, -1};
        Queue <Integer> qx = new LinkedList<Integer>();
        Queue <Integer> qy = new LinkedList<Integer>();
        qx.offer(x);
        qy.offer(y);
        visited[x][y] = true;

        while (!qx.isEmpty() && !qy.isEmpty()) {
            int cx = qx.poll();
            int cy = qy.poll();
            // iterate over up, left, right, down 4 direction
            for (int k = 0; k < 4; k++) {
                int nx = cx + dx[k];
                int ny = cy + dy[k];
                // if within boundaries and not visited and is a part of island
                if (nx >= 0 && nx < m && ny >= 0 && ny < n && !visited[nx][ny] && grid[nx][ny] == '1') {
                    visited[nx][ny] = true; 
                    qx.offer(nx);
                    qy.offer(ny);
                }
            }
        }
    }
}

Union Find (Verbose, 19 ms AC)

class UnionFind {
    public int[] parent;
    public int[] size;

    // Initialize UnionFind
    public UnionFind() {}

    public UnionFind(int n) {
        this.parent = new int[n];
        size = new int[n];
        for (int i = 0; i < n; i++) {
            this.parent[i] = i;
            size[i] = 1;
        }
    }

    // Find Method
    public int find(int x) {
        return compressedFind(x);
    }

    // Find Method with Path Compression
    public int compressedFind(int x) {
        while (x != parent[x]) {
            parent[x] = parent[parent[x]];
            x = parent[x];
        }
        return x;
    }

    // Union Method with Weight
    public void union(int x, int y) {
        int parentX = find(x);
        int parentY = find(y);
        if (parentX == parentY) {
            return;
        }

        if (this.size[parentX] < this.size[parentY]) {
            this.size[parentY] += this.size[parentX];
            this.parent[parentX] = parentY;
        } else {
            this.size[parentX] += this.size[parentY];
            this.parent[parentY] = parentX;
        }
    }
}

class IslandUnionFind extends UnionFind {
    public int count;

    public void initIslands(boolean grid[][]) {
        count = 0;
        int m = grid.length;
        int n = grid[0].length;
        this.parent = new int[m * n];
        this.size = new int[m * n];

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == true) {
                    parent[i * n + j] = i * n + j;
                    count++;
                } else {
                    parent[i * n + j] = -1;
                }
                this.size[i * n + j] = 1;
            }
        }
    }

    public void updateCount(int k) {
        count = count + k;
    }

    public int getCount() {
        return count;
    }
}

public class Solution {

    public boolean insideGrid(int x, int y, int m, int n) {
        return (x >= 0 && y >= 0 && x < m && y < n);
    }
    /**
     * @param grid a boolean 2D matrix
     * @return an integer
     */
    public int numIslands(boolean[][] grid) {
        // Check Input
        if(grid==null || grid.length==0 || grid[0].length==0) {
            return 0;
        }

        IslandUnionFind uf = new IslandUnionFind();
        uf.initIslands(grid);

        int m = grid.length;
        int n = grid[0].length;

        // Helper Direction Array
        int[] dx = new int[] {-1, 1, 0, 0};
        int[] dy = new int[] {0, 0, -1, 1};

        // Row and Column Traversal
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (insideGrid(i, j, m, n) && grid[i][j]) {

                    // 4 directions for each point
                    for (int k = 0; k < 4; k++) {
                        int nx = i + dx[k];
                        int ny = j + dy[k];
                        if (insideGrid(nx, ny, m, n) && grid[nx][ny]) {
                            int cparent = uf.find(i * n + j);
                            int nparent = uf.find(nx * n + ny);
                            if (cparent != nparent) {
                                uf.union(cparent, nparent);
                                uf.updateCount(-1);
                            }
                        }
                    }
                }
            }
        }

        return uf.getCount();
    }
}

Another Union Find (8ms AC)

class Solution {
    int[][] distance = { { 1, 0 }, {-1, 0 }, { 0, 1 }, { 0, -1 } };
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        UnionFind uf = new UnionFind(grid);
        int rows = grid.length;
        int cols = grid[0].length;
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == '1') {
                    for (int[] d: distance) {
                        int x = i + d[0];
                        int y = j + d[1];
                        if (x >= 0 && x < rows && y >= 0 && y < cols && grid[x][y] == '1') {
                            int id1 = i * cols + j;
                            int id2 = x * cols + y;
                            uf.union(id1, id2);
                        }
                    }
                }
            }
        }
        return uf.count;
    }

    class UnionFind {
        int[] father;
        int m, n;
        int count = 0;
        UnionFind(char[][] grid) {
            m = grid.length;
            n = grid[0].length;
            father = new int[m * n];
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    if (grid[i][j] == '1') {
                        int id = i * n + j;
                        father[id] = id;
                        count++;
                    }
                }
            }
        }
        public void union(int node1, int node2) {
            int find1 = find(node1);
            int find2 = find(node2);
            if (find1 != find2) {
                father[find1] = find2;
                count--;
            }
        }
        public int find(int node) {
            if (father[node] == node) {
                return node;
            }
            father[node] = find(father[node]);
            return father[node];
        }
    }
}

Solution

• programcreek: LeetCode – Number of Islands (Java)
• LeetCode Discussion: AC Java Solution using Union-Find with explanations