Leetcode - Number Of Islands

岛屿数量计算

最新推荐文章于 2022-09-12 16:34:28 发布

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本文介绍了一种使用广度优先搜索(BFS)算法计算二维网格中岛屿数量的方法。通过遍历网格，遇到陆地('1')时计数加一，并启动BFS搜索所有相邻的陆地，将其标记以防重复计数。

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:

11110
11010
11000
00000
Answer: 1

Example 2:

11000
11000
00100
00011
Answer: 3

[分析] 思路和Surrounded Region神似，遍历grid，遇到1计数加1,且从该位置处开始进行BFS搜索相连的 1 ，将BFS搜索路径上的 1 全部标记为非 0 非 1 字符表明已搜索，grid遍历完得到的计数即为所求。


public class Solution {
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0)
            return 0;
        int rows = grid.length, cols = grid[0].length;
        int num = 0;
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == '1') {
                    bfs(i, j, grid);
                    num++;
                }
            }
        }
        return num;
    }
    private void bfs(int i, int j, char[][] grid) {
        grid[i][j] = '2';
        LinkedList<Integer> queue = new LinkedList<Integer>();
        int cols = grid[0].length;
        queue.offer(i * cols + j);
        while (!queue.isEmpty()) {
            int idx = queue.poll();
            int r = idx / cols;
            int c = idx % cols;
            visit(r + 1, c, grid, queue);
            visit(r - 1, c, grid, queue);
            visit(r, c + 1, grid, queue);
            visit(r, c - 1, grid, queue);
        }
    }
    private void visit(int i, int j, char[][] grid, LinkedList<Integer> queue) {
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] != '1')
            return;
        grid[i][j] = '2';
        queue.offer(i * grid[0].length + j);
    }
}