spring自定义注解

200. 岛屿数量

难度中等1136收藏分享切换为英文接收动态反馈

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'

题解

class Solution {
    int[][] direction = {{-1,0}, {1,0}, {0,1}, {0,-1}};

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

    public void bfs(char[][] grid, int i, int j) {
        if (!isInGrid(grid, i, j)) return;
        if (grid[i][j] != '1') return ;
        grid[i][j] = '2';
        for(int[] dir : direction) {
            bfs(grid, i + dir[0], j + dir[1]);
        }
    }

    boolean isInGrid(char[][] grid, int i ,int j){
        return i >= 0 && i < grid.length && j >= 0 && j < grid[0].length;
    }
}

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