算法-岛屿数量

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。

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

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

代码实现：

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;
    }

    private void dfs(char[][] grid, int r, int c) {
        if (!isInGrid(grid, r, c)) {
            return;
        }
        // if (grid[r][c] == '2') {
        //     return;
        // }
        if (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);
    }

    private boolean isInGrid(char[][] grid, int r, int c) {
        return r >= 0 && c >= 0 && r < grid.length && c < grid[0].length;
    }
}

原题链接：力扣

视频讲解：岛屿数量_哔哩哔哩_bilibili