LeetCode：200. 岛屿数量（BFS Java）-优快云博客

本文链接：https://blog.youkuaiyun.com/Cosmoshhhyyy/article/details/137789888

本文介绍了一个编程问题，要求计算由1（陆地）和0（水）组成的二维网格中的岛屿数量。使用了广度优先搜索(BFS)算法进行遍历，避免了重复计数，确保每个岛屿只被计算一次。

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200. 岛屿数量

题目描述：

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'

实现代码与解析：

BFS

class Solution {
    int res = 0;
    int n;
    int m;
    int[] dx = new int[]{0, 1, 0, -1};
    int[] dy = new int[]{-1, 0, 1, 0};

    public int numIslands(char[][] grid) {

        n = grid.length;
        m = grid[0].length;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == '0') continue;
                bfs(grid, i, j);
                res++;
            }
        }
        return res;
    }

    public void bfs(char[][] g, int x, int y) {

        Queue<Pair<Integer, Integer>> q = new LinkedList<>();
        q.offer(new Pair<>(x, y));
        g[x][y] = '0';
        while (!q.isEmpty()) {
            Pair<Integer, Integer> t = q.peek();
            q.poll();
            x = t.getKey();
            y = t.getValue();
            for (int i = 0; i < 4; i++) {
                int xt = x + dx[i];
                int yt = y + dy[i];
                if (xt < 0 || xt > n - 1 || yt < 0 || yt > m - 1) continue;
                if (g[xt][yt] == '0') continue;
                g[xt][yt] = '0';
                q.offer(new Pair<>(xt, yt));
            }
        }

    }
}