[LeetCode]Number of Islands

最新推荐文章于 2025-09-10 15:24:29 发布

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最新推荐文章于 2025-09-10 15:24:29 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/dick159/p/5060940.html

本文通过DFS算法解决岛屿数量计算问题，清晰地阐述了如何将连通子图统一集合化并计数。

题目:

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.

Exp:

Input:

11110

11010

11000

00000

OutPut:

Input:

11000

00100

00011

Output:

题目的意思就是很明确，可以转化为求连通子图的数，这里我先用dfs递归求解。

首先，明确这是个求连通子图的题,只要将连起来的子图都统一集合化，说白了就是把连起来的1看成一个单位，所以要把他变成统一的区分于其他的标志，这里就先弄成2。

So,上代码！

public class Solution {

    int ret = 0;
    int m;
    int n;

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

        return ret;
    }

    public void dfs(char[][] grid, int x, int y) {
        if (x < 0 || y < 0 || x >= m || y >= n)
            return;
        else {
            if (grid[x][y] == '1') {
                grid[x][y] = '2';
                dfs(grid, x - 1, y);
                dfs(grid, x + 1, y);
                dfs(grid, x, y + 1);
                dfs(grid, x, y - 1);
            }
        }
    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Solution s = new Solution();
        Scanner sc = new Scanner(System.in);
        char[][] grid = new char[5][5];
        for (int i = 0; i < 4; i++) {
            String str = sc.next();
            grid[i] = str.toCharArray();
        }
        System.out.println(s.numIslands(grid));
    }

}