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。

样例描述

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

思路

FloodFill算法 DFS

代码

class Solution {
    
    
    //方向数组
    int dx[] = new int[]{
    
    1, 0, -1 ,0};
    int dy[] = new int[]{
    
    0, 1, 0, -1};
    public int numIslands(char[][] grid) {
    
    
        int res = 0;
        for (int i = 0; i < grid.length; i ++ ) {
    
    
            for (int j = 0; j < grid[i].length; j ++ ) {
    
    
                //是岛屿就dfs遍历
                if (grid[i][j] == '1') {
    
    
                    dfs(grid, i, j);
                    res ++;
                }
            }
        }
        return res;
    }
    public void dfs(char[][] grid, int x, int y) {
    
    
        //遍历过后就标记成0，变成水
        grid[x][y] = '0';
        for (int i = 0; i < 4; i ++ ) {
    
    
            int a = x + dx[i], b = y + dy[i];
            //没有越界并且为陆地时，继续遍历周围
            if (a >= 0 && a < grid.length && b >= 0 && b < grid[0].length && grid[a][b] == '1') {
    
    
                 dfs(grid, a, b);
            }
        }
    }
}