200. Number of Islands

DFS

class Solution {
    private static final int[][] DIRECTIONS = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

    public int numIslands(char[][] grid) {
        int m = grid.length, n = grid[0].length;

        int landCount = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    dfs(grid, m, n, i, j);
                    landCount++;
                }
            }
        }

        return landCount;
    }

    private void dfs(char[][] grid, int m, int n, int i, int j) {
        grid[i][j] = '0';
        for (int[] direction : DIRECTIONS) {
            int nextI = i + direction[0];
            int nextJ = j + direction[1];
            if (nextI >= 0 && nextI < m && nextJ >= 0 && nextJ < n && grid[nextI][nextJ] == '1') {
                dfs(grid, m, n, nextI, nextJ);
            }
        }
    }
}

BFS

class Solution {
    private static final int[][] DIRECTIONS = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

    public int numIslands(char[][] grid) {
        int m = grid.length, n = grid[0].length;

        int landCount = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    bfs(grid, m, n, i, j);
                    landCount++;
                }
            }
        }

        return landCount;
    }

    private void bfs(char[][] grid, int m, int n, int i, int j) {
        Queue<int[]> queue = new LinkedList<>(); // 队列中的点为已经清除了陆地状态的点
        grid[i][j] = '0'; // 注意在入队前标记以防止重复入队
        queue.offer(new int[]{i, j});

        while (!queue.isEmpty()) {
            int[] point = queue.poll();
            
            for (int[] direction : DIRECTIONS) {
                int nextI = point[0] + direction[0];
                int nextJ = point[1] + direction[1];

                if (nextI < 0 || nextI >= m || nextJ < 0 || nextJ >= n || grid[nextI][nextJ] == '0') {
                    continue;
                }

                grid[nextI][nextJ] = '0'; // 注意在入队前标记以防止重复入队
                queue.offer(new int[]{nextI, nextJ});
            }
        }
    }
}

注意入队前标记。

UnionFind

class Solution {
    private static class UnionFind {

        private final int[] parent;
        private int areaCount;

        public UnionFind(int count) {
            this.parent = new int[count];
            // 不要忘记初始化每个索引指向自己
            for (int i = 0; i < count; i++) {
                this.parent[i] = i;
            }
            this.areaCount = count;
        }

        public void union(int x, int y) {
            int xRoot = getRoot(x);
            int yRoot = getRoot(y);
            if (xRoot == yRoot) {
                return;
            }

            parent[xRoot] = yRoot; // 注意此处是把 x 的根节点指向 y 的根节点，而不能把 x 指向 y 的根节点，否则 x 的父节点等的指向未被改变
            areaCount--;
        }

        public int getRoot(int x) {
            while (x != parent[x]) {
                parent[x] = parent[parent[x]]; // path compression
                x = parent[x];
            }
            return x;
        }

    }

    public int numIslands(char[][] grid) {
        int m = grid.length, n = grid[0].length;

        int[][] directions = new int[][]{{1, 0}, {0, 1}};
        UnionFind unionFind = new UnionFind(m * n);

        int water = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    for (int[] direction : directions) {
                        int nextI = i + direction[0];
                        int nextJ = j + direction[1];
                        if (nextI < m && nextJ < n && grid[nextI][nextJ] == '1') {
                            unionFind.union(getIndex(n, i, j), getIndex(n, nextI, nextJ));
                        }
                    }
                } else {
                    water++;
                }
            }
        }

        return unionFind.areaCount - water;
    }

    private int getIndex(int width, int i, int j) {
        return i * width + j;
    }
}

Reference

200. Number of Islands