LeetCode 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'

思路

dfs搜索，每次遇到一个岛屿则将陆地全部变成海水，记录遇到的岛屿数即可。

class Solution {

    public int[][] directions = { {1, 0}, {-1, 0}, {0, 1}, {0, -1} };

    public char[][] grid = null;

    int m = 0, n = 0;

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

    void dfs(int x, int y) {
        if (grid[x][y] == '0') return;
        grid[x][y] = '0';
        for (int d = 0; d < 4; d++) {
            int newX = directions[d][0] + x;
            int newY = directions[d][1] + y;

            if (0 <= newX && newX < m && 0 <= newY && newY < n)
                dfs(newX, newY);
        }
    }
}

复杂度

• 时间复杂度：
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