【宽度优先搜索 BFS】day54_200. 岛屿数量

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：

思路：dfs

时间复杂度：O(mn)

空间复杂度：O(mn)

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

    public void dfs(char[][] grid,int i, int j){
        //防止 i 和 j 越界，也就是防止超出岛屿（上下左右）的范围。特别注意当遍历到海洋的时候也退出循环
        if(i<0||j<0||i>=grid.length||j>=grid[0].length||grid[i][j]=='0') return;
        //将遍历过的陆地改为海洋，防止重复遍历
        grid[i][j]='0';
        //上
        dfs(grid,i+1,j);
        //下
        dfs(grid,i-1,j);
        //右
        dfs(grid,i,j+1);
        //左
        dfs(grid,i,j-1);
    }
}