岛屿数量

2020-12-06

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

示例 2：

输入：grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
输出：3
 

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'
通过次数183,225提交次数356,708

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。

深度优先遍历，将遍历过的元素置为非’0’和’1’的字符,统计所有入口的数量

class Solution {
    private static int[][] road = new int[][]{{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
    public int numIslands(char[][] grid) {
        int count = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == '1') {
                    count++;
                    deepRoute(grid, i, j);
                }
            }
        }
        return count;
    }

    public void deepRoute(char[][] grid, int x, int y) {
        ArrayDeque<int[]> stack = new ArrayDeque<>();
        stack.push(new int[]{x, y});
        grid[x][y] = '2';
        while (!stack.isEmpty()) {
            int[] location = stack.pop();
            for (int i = 0; i < 4; i++) {
                int x_ = location[0] + road[i][0];
                int y_ = location[1] + road[i][1];
                if (x_ >= 0 && x_ < grid.length && y_ >= 0 && y_ < grid[0].length) {
                    if (grid[x_][y_] == '1') {
                        stack.push(new int[]{x_, y_});
                        grid[x_][y_] = '2';
                    }
                }
            }
        }

    }
}

深度优先遍历，但是效率更高

class Solution {
    void dfs(char[][] grid, int r, int c) {
        int nr = grid.length;
        int nc = grid[0].length;

        if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == '0') {
            return;
        }

        grid[r][c] = '0';
        dfs(grid, r - 1, c);
        dfs(grid, r + 1, c);
        dfs(grid, r, c - 1);
        dfs(grid, r, c + 1);
    }

    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;
        for (int r = 0; r < nr; ++r) {
            for (int c = 0; c < nc; ++c) {
                if (grid[r][c] == '1') {
                    ++num_islands;
                    dfs(grid, r, c);
                }
            }
        }

        return num_islands;
    }
}

广度优先遍历

class Solution {
    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;

        for (int r = 0; r < nr; ++r) {
            for (int c = 0; c < nc; ++c) {
                if (grid[r][c] == '1') {
                    ++num_islands;
                    grid[r][c] = '0';
                    Queue<Integer> neighbors = new LinkedList<>();
                    neighbors.add(r * nc + c);
                    while (!neighbors.isEmpty()) {
                        int id = neighbors.remove();
                        int row = id / nc;
                        int col = id % nc;
                        if (row - 1 >= 0 && grid[row-1][col] == '1') {
                            neighbors.add((row-1) * nc + col);
                            grid[row-1][col] = '0';
                        }
                        if (row + 1 < nr && grid[row+1][col] == '1') {
                            neighbors.add((row+1) * nc + col);
                            grid[row+1][col] = '0';
                        }
                        if (col - 1 >= 0 && grid[row][col-1] == '1') {
                            neighbors.add(row * nc + col-1);
                            grid[row][col-1] = '0';
                        }
                        if (col + 1 < nc && grid[row][col+1] == '1') {
                            neighbors.add(row * nc + col+1);
                            grid[row][col+1] = '0';
                        }
                    }
                }
            }
        }

        return num_islands;
    }
}

并查集

同样地，我们也可以使用并查集代替搜索。

为了求出岛屿的数量，我们可以扫描整个二维网格。如果一个位置为 1，则将其与相邻四个方向上的 1 在并查集中进行合并。

最终岛屿的数量就是并查集中连通分量的数目。


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] = '0';
                    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();
    }
}

时间复杂度：O(MN∗α(MN))，其中 M 和 N 分别为行数和列数。注意当使用路径压缩（见 find 函数）和按秩合并（见数组 rank）实现并查集时，单次操作的时间复杂度为α(MN)，其中α(x) 为反阿克曼函数，当自变量 x 的值在人类可观测的范围内（宇宙中粒子的数量）时，函数 α(x) 的值不会超过 5，因此也可以看成是常数时间复杂度。

空间复杂度：O(MN)，这是并查集需要使用的空间。

参考文献

作者：LeetCode

链接：https://leetcode-cn.com/problems/number-of-islands/solution/dao-yu-shu-liang-by-leetcode/

来源：力扣（LeetCode）

著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。