LeetCode in Python 695. Max Area of Island
内容导读
互联网集市收集整理的这篇技术教程文章主要介绍了LeetCode in Python 695. Max Area of Island,小编现在分享给大家,供广大互联网技能从业者学习和参考。文章包含1334字,纯文字阅读大概需要2分钟。
内容图文
![LeetCode in Python 695. Max Area of Island](/upload/InfoBanner/zyjiaocheng/764/8ac1a4cabf244d7a9b8b73d090902ed1.jpg)
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.
class Solution(object): def maxAreaOfIsland(self, grid): h, w = len(grid), len(grid[0]) def dfs(x, y): if 0 <= x < h and 0 <= y < w and grid[x][y]: grid[x][y] = 0 return 1 + dfs(x-1, y) + dfs(x+1, y) + dfs(x, y-1) + dfs(x, y+1) return 0 areas = [dfs(i, j) for i in range(h) for j in range(w) if gird[i][j]] return max(areas) if areas else 0
四方向dfs遍历,注意每次进入递归函数且确定是1后要置0,这样对于每一个island来说,第一次找到属于它的1时其它的全部都找到了,把他们标记为0后避免了重复查找。用for(h,w)在图中搜索尚未访问的1即可找到最大值。
内容总结
以上是互联网集市为您收集整理的LeetCode in Python 695. Max Area of Island全部内容,希望文章能够帮你解决LeetCode in Python 695. Max Area of Island所遇到的程序开发问题。 如果觉得互联网集市技术教程内容还不错,欢迎将互联网集市网站推荐给程序员好友。
内容备注
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 gblab@vip.qq.com 举报,一经查实,本站将立刻删除。
内容手机端
扫描二维码推送至手机访问。