Leetcode算法--岛屿数量_算法岛屿数量-优快云博客

本文链接：https://blog.youkuaiyun.com/j18874964028sss/article/details/132134532

文章讨论了如何通过编程解决一个计算给定二维网格中岛屿数量的问题，利用深度优先搜索（DFS）遍历并标记陆地区域，避免重复计数。代码给出了一个名为Solution的类，包含numIslands、dfs和isInGrid方法来实现这一功能。

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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