public int solve (char[][] grid) {
        // write code here

        int count=0;    //返回结果

        for(int i=0;i<grid.length;i++){
            for(int j=0;j<grid[i].length;j++){
                //从第一个开始判断,如果是1进行dfs并count++
                if(grid[i][j]=='1'){
                    dfs(grid,i,j);
                    count++;
                }
            }
        }
        return count;
    }

    //进行dfs查找
    public void dfs(char[][] grid,int i,int j){
        //边界限制,以及遇到海洋就终止
        if(i<0||i>=grid.length || j<0|| j>=grid[0].length || grid[i][j]=='0') return;

        //已经探查过的陆地就使之为“海洋”,避免出现 栈溢出
        grid[i][j]='0';
        //上下左右
        dfs(grid,i+1,j);
        dfs(grid,i-1,j);
        dfs(grid,i,j+1);
        dfs(grid,i,j-1);
    }