给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

 

示例 1：

输入：grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
输出：1

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-islands
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

解析

很明显套公式就好了。

解法二我们练习一下并查集的解法。

解法一

class Solution {
    public int numIslands(char[][] grid) {
        int res = 0;
        for (int r = 0; r < grid.length; r++) {
            for (int c = 0; c < grid[0].length; c++) {
                if (grid[r][c] == '1') {
                    dfs(grid, r, c);
                    res++;
                }
            }
        }
        return res;
    }

    void dfs(char[][] grid, int r, int c) {
        // 函数因为「坐标 (r, c) 超出网格范围」
        // 函数因为「当前格子是海洋格子」或者【被访问过】
        if (!inArea(grid, r, c) || grid[r][c] != '1') {
            return;
        }
        grid[r][c] = '2';
        dfs(grid, r - 1, c);
        dfs(grid, r + 1, c);
        dfs(grid, r, c - 1);
        dfs(grid, r, c + 1);
    }

    // 判断坐标 (r, c) 是否在网格中
    boolean inArea(char[][] grid, int r, int c) {
        return 0 <= r && r < grid.length
                && 0 <= c && c < grid[0].length;
    }
}

解法二

并查集,我们对于并查集的count设置为1的个数，每一次遍历到1就进行将他的上下左右都进行union

注意还有另外一种并查集的写法，就是将二维数组的x*y 作为总的连通分量，最后用uf.count()-space空格数就得到了答案。

class Solution {
    class UnionFind {
        int count;
        int[] parent;
        int[] rank;

        public UnionFind(char[][] grid) {
            count = 0;
            int m = grid.length;
            int n = grid[0].length;
            parent = new int[m * n];
            rank = new int[m * n];
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (grid[i][j] == '1') {
                        parent[i * n + j] = i * n + j;
                        ++count;
                    }
                    rank[i * n + j] = 0;
                }
            }
        }

        public int find(int i) {
            if (parent[i] != i) parent[i] = find(parent[i]);
            return parent[i];
        }

        public void union(int x, int y) {
            int rootx = find(x);
            int rooty = find(y);
            if (rootx != rooty) {
                if (rank[rootx] > rank[rooty]) {
                    parent[rooty] = rootx;
                } else if (rank[rootx] < rank[rooty]) {
                    parent[rootx] = rooty;
                } else {
                    parent[rooty] = rootx;
                    rank[rootx] += 1;
                }
                --count;
            }
        }

        public int getCount() {
            return count;
        }
    }

    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }

        int nr = grid.length;
        int nc = grid[0].length;
        int num_islands = 0;
        UnionFind uf = new UnionFind(grid);
        for (int r = 0; r < nr; ++r) {
            for (int c = 0; c < nc; ++c) {
                if (grid[r][c] == '1') {
                    grid[r][c] = '2';
                    if (r - 1 >= 0 && grid[r-1][c] == '1') {
                        uf.union(r * nc + c, (r-1) * nc + c);
                    }
                    if (r + 1 < nr && grid[r+1][c] == '1') {
                        uf.union(r * nc + c, (r+1) * nc + c);
                    }
                    if (c - 1 >= 0 && grid[r][c-1] == '1') {
                        uf.union(r * nc + c, r * nc + c - 1);
                    }
                    if (c + 1 < nc && grid[r][c+1] == '1') {
                        uf.union(r * nc + c, r * nc + c + 1);
                    }
                }
            }
        }

        return uf.getCount();
    }
}

//另外一种写法
public class Solution {

    private int rows;
    private int cols;

    public int numIslands(char[][] grid) {
        rows = grid.length;
        if (rows == 0) {
            return 0;
        }
        cols = grid[0].length;

        // 空地的数量
        int spaces = 0;
        UnionFind unionFind = new UnionFind(rows * cols);
        int[][] directions = {{1, 0}, {0, 1},{-1, 0}, {0, -1}};
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == '0') {
                    spaces++;
                } else {
                    // 此时 grid[i][j] == '1'
                    for (int[] direction : directions) {
                        int newX = i + direction[0];
                        int newY = j + direction[1];
                        // 先判断坐标合法，再检查右边一格和下边一格是否是陆地
                        if (newX >=0 && newY>=0 && newX < rows && newY < cols && grid[newX][newY] == '1') {
                            unionFind.union(getIndex(i, j), getIndex(newX, newY));
                        }
                    }
                }
            }
        }
        return unionFind.getCount() - spaces;
    }

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

    private class UnionFind {
        /**
         * 连通分量的个数
         */
        private int count;
        private int[] parent;

        public int getCount() {
            return count;
        }

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

        private int find(int x) {
            while (x != parent[x]) {
                parent[x] = parent[parent[x]];
                x = parent[x];
            }
            return x;
        }

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

            parent[xRoot] = yRoot;
            count--;
        }
    }
}

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