[leetcode] Max Area of Island

Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

[[0,0,1,0,0,0,0,1,0,0,0,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,1,1,0,1,0,0,0,0,0,0,0,0],
 [0,1,0,0,1,1,0,0,1,0,1,0,0],
 [0,1,0,0,1,1,0,0,1,1,1,0,0],
 [0,0,0,0,0,0,0,0,0,0,1,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,0,0,0,0,0,0,1,1,0,0,0,0]]
Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

[[0,0,0,0,0,0,0,0]]
Given the above grid, return 0.

Note: The length of each dimension in the given grid does not exceed 50.


分析:题目翻译一下:要求计算连续1的个数。看题目也是一个DFS问题,当遇到一个1,然后就从上、下、左、又四个方向去搜索。这里要注意搜索边界,如果在边界地方还要搜索,有可能数组越界。
比如
现在加入搜索到第二行第二列的1,然后向上是0,向下是1,再往下还是1,再往下是0。这里要注意如果是1的话,要用一个visited数组来保存访问记录。可以降低重复访问次数。
代码如下:
 1 public class maxareaofisland {
 2     int maxarea = 0;
 3     public int maxAreaOfIsland(int[][] grid) {
 4         boolean[][] visited = new boolean[grid.length][grid[0].length];
 5         for ( int i = 0 ; i < grid.length ; i ++ ){
 6             for ( int j = 0 ; j < grid[i].length ; j ++ ){
 7                 int temparea = helper(i,j,grid,visited);
 8                 maxarea = Math.max(maxarea,temparea);
 9             }
10         }
11         return maxarea;
12     }
13 
14     private int helper(int i, int j, int[][] grid, boolean[][] visited) {
15         if (  i < 0 || j < 0 || i > grid.length - 1 || j > grid.length - 1) return 0;
16         if ( grid[i][j] == 0 || visited[i][j]) return 0;
17         visited[i][j] = true;
18         return 1 + helper(i+1,j,grid,visited)+helper(i-1,j,grid,visited)+helper(i,j+1,grid,visited)+helper(i,j-1,grid,visited);
19     }
20 }

    运行时间23ms,击败36.30%的提交。

    看了一下其他的方法,基本都是深搜。只是一些技巧问题,就不管了。

转载于:https://www.cnblogs.com/boris1221/p/9367404.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值