码迷,mamicode.com
首页 > 其他好文 > 详细

[LeetCode] 695. Max Area of Island_Easy tag: DFS/BFS

时间:2018-08-08 00:57:50      阅读:137      评论:0      收藏:0      [点我收藏+]

标签:maximum   bar   max   for   note   idea   nbsp   grid   sid   

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.

 
这个题目思路就是BFS/DFS.
 
1. Constraints
1) edge case len(grid), len(grid[0]) == 0 => 0
2)element will be 1, or 0
 
2. Ideas
DFS    T: O(m*n)     S; O(m*n)
 
3. Code
class Solution:
    def maxAreaLand(self, grid):
        if not grid or len(grid[0]) == 0: return 0
        lr, lc, ans, visited = len(grid), len(grid[0]), 0, set()
        def dfs(r,c):
            if 0 <= r <len(grid) and 0 <= c < len(grid[0]) and grid[r][c] and (r,c) not in visted:
                 return 1 + dfs(r-1, c) + dfs(r+1, c) + dfs(r, c-1) + dfs(r, c+1)
            return 0
        for i in range(lr):
            for j in range(lc):
                ans = max(ans, dfs(i,j))
        return ans

 


 
 
 

[LeetCode] 695. Max Area of Island_Easy tag: DFS/BFS

标签:maximum   bar   max   for   note   idea   nbsp   grid   sid   

原文地址:https://www.cnblogs.com/Johnsonxiong/p/9440473.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!