leetcode 695. 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.
解法1:dfs
class Solution(object): def maxAreaOfIsland(self, grid): """ :type grid: List[List[int]] :rtype: int """ # reset 1 to 0 use dfs, and return the ones numer row = len(grid) col = len(grid[0]) ans = 0 def dfs(grid, i, j): if i==row or j==col or i<0 or j<0 or grid[i][j]==0: return 0 res = 1 grid[i][j] = 0 res += dfs(grid, i+1, j) res += dfs(grid, i-1, j) res += dfs(grid, i, j-1) res += dfs(grid, i, j+1) return res for i in range(0, row): for j in range(0, col): if grid[i][j] == 1: ans = max(ans, dfs(grid, i, j)) return ans
解法2:BFS,必须在if里将1reset为0,否则有重复:
class Solution(object): def maxAreaOfIsland(self, grid): """ :type grid: List[List[int]] :rtype: int """ # reset 1 to 0 use bfs, and return the ones numer row = len(grid) col = len(grid[0]) ans = 0 def bfs(grid, i, j): assert grid[i][j] == 1 q = [(i,j)] grid[i][j]=0 res = 1 while q: q2 = [] for i,j in q: if i+1<row and grid[i+1][j]==1: grid[i+1][j]=0 q2.append((i+1,j)) res += 1 if i-1>=0 and grid[i-1][j]==1: grid[i-1][j]=0 q2.append((i-1,j)) res += 1 if j+1<col and grid[i][j+1]==1: grid[i][j+1]=0 q2.append((i,j+1)) res += 1 if j-1>=0 and grid[i][j-1]==1: grid[i][j-1]=0 q2.append((i,j-1)) res += 1 q = q2 return res for i in range(0, row): for j in range(0, col): if grid[i][j] == 1: ans = max(ans, bfs(grid, i, j)) return ans
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析
· 没有源码,如何修改代码逻辑?
· 全程不用写代码,我用AI程序员写了一个飞机大战
· MongoDB 8.0这个新功能碉堡了,比商业数据库还牛
· 记一次.NET内存居高不下排查解决与启示
· 白话解读 Dapr 1.15:你的「微服务管家」又秀新绝活了
· DeepSeek 开源周回顾「GitHub 热点速览」
2017-03-17 MongoDB C++ gridfs worked example
2017-03-17 linux编程 fmemopen函数打开一个内存流 使用FILE指针进行读写访问