图：岛屿数量

leetcode：200. 岛屿数量

题目描述

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

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

解题思路

每一次遇到陆地 将陆地1遍历 变为海水0。

DFS

沿着一个方向遍历，直到遇到海水。

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

    public int numIslands(char[][] grid) {
        int res = 0;
        int m = grid.length, n = grid[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    res++;
                    dfs(grid, i, j);
                }
            }
        }
        return res;
    }

    void dfs(char[][] grid, int row, int col) {
        if (row < 0 || row >= grid.length || 
            col < 0 || col >= grid[0].length || 
            grid[row][col] == '0') {
                return;
        }

        grid[row][col] = '0';
        
        for (int[] dir : DIRS) {
            dfs(grid, row + dir[0], col + dir[1]);
        }
    }
}

BFS

一圈一圈，队列存储当前 陆地 周围的陆地。

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

    public int numIslands(char[][] grid) {
        int res = 0;
        int m = grid.length, n = grid[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    res++;
                    bfs(grid, i, j);
                }
            }
        }
        return res;
    }

    void bfs(char[][] grid, int row, int col) {
        int m = grid.length, n = grid[0].length;
        Deque<int[]> queue = new ArrayDeque<>();
        queue.addLast(new int[]{row, col});
        grid[row][col] = '0';
        while (!queue.isEmpty()) {
            int[] curr = queue.removeFirst();
            int currRow = curr[0], currCol = curr[1];
            for (int[] dir : DIRS) {
                int nextRow = currRow + dir[0], nextCol = currCol + dir[1];
                if (nextRow >= 0 && nextRow < m &&
                    nextCol >= 0 && nextCol < n &&
                    grid[nextRow][nextCol] == '1') {
                        queue.addLast(new int[]{nextRow, nextCol});
                        grid[nextRow][nextCol] = '0';
                    }
            }
        }
    }
}