Medium | LeetCode 200. 岛屿数量 | 矩阵 + DFS

200. 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。

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'

解题思路

经典的矩阵深度优先搜索问题

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;
}

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;
    }
    // 没访问过一次, 就将其标记为0, 表示不可访问。
    grid[r][c] = '0';
    dfs(grid, r - 1, c);
    dfs(grid, r + 1, c);
    dfs(grid, r, c - 1);
    dfs(grid, r, c + 1);
}