200. Number of Islands

最新推荐文章于 2021-11-25 12:03:14 发布

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本文介绍了一种使用深度优先搜索算法解决经典岛屿数量计数问题的方法。通过遍历二维网格地图，识别由'1'表示的陆地，并计算彼此相连的陆地形成岛屿的数量。代码示例展示了如何实现这一算法。

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Input:
11110
11010
11000
00000

Output: 1
Example 2:

Input:
11000
11000
00100
00011

Output: 3

本题是一个非常经典的深度优先搜索的题目（代码我都能背出来了）。代码如下：

    int[] dx = new int[]{0, 1, 0, -1};
    int[] dy = new int[]{1, 0, -1, 0};

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

    private void dfs(int i, int j, char[][] grid) {
        for(int k = 0; k < 4; ++k) {
            int x = i + dx[k];
            int y = j + dy[k];
            if(0 <= x && x < grid.length && 0 <= y && y < grid[0].length && grid[x][y] == '1') {
                grid[x][y] = '0';
                dfs(x, y, grid);
            }
        }
    }