算法之统计岛屿数量

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。
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

class Solution {    public int numIslands(char[][] grid) {        int count = 0;        int height = grid.length;        int w = grid[0].length;        for(int i=0; i < height;i++ ) {            for(int j = 0; j < w; j++) {                if(grid[i][j] == '1') {                    dfs(grid, i, j);                    count++;                }            }        }        return count;    }
    public void dfs(char[][] grid, int i, int j) {        if(i >= grid.length || i < 0 ) return;        if(j < 0 || j >= grid[i].length) return;        if(grid[i][j] == '1') {            grid[i][j] = '2';            //print(grid); //可以开启打印            dfs(grid, i+1, j);            dfs(grid, i, j+1);            dfs(grid, i-1, j);             dfs(grid, i, j - 1);        }    }
    //打印每个处理过程，标记后的数组效果    private void print(char[][] grid) {        for(int i=0;i<grid.length;i++) {            for(int j=0; j < grid.length; j++) {                System.out.print(grid[i][j] + " ");            }            System.out.println("");        }        System.out.println("---------------------");    }}

        //我们一般从左到由，从上到下进行处理扫描，但是存在一种需要反向处理上一行的情况        //2 0 1       2 0 1     2 0 2        //2 0 1  -->  2 0 2 --> 2 0 2        //2 2 2       2 2 2     2 2 2

        dfs(grid, i-1, j);         dfs(grid, i, j - 1);