[LeetCode]Max Area of Island

题目描述:

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.

题目大意:

给定二维格子grid,上下左右相邻的1组成岛屿,求岛屿的最大面积。

解题思路:

FloodFill(洪水填充)

Python代码:

class Solution(object):
    def maxAreaOfIsland(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        ans = 0
        for x in range(len(grid)):
            for y in range(len(grid[0])):
                if grid[x][y]:
                    ans = max(ans, self.bfs(grid, x, y))
        return ans
    def bfs(self, grid, x, y):
        dxs = [1, 0, -1, 0]
        dys = [0, 1, 0, -1]
        queue = [(x, y)]
        grid[x][y] = 0
        ans = 0
        while queue:
            x, y = queue.pop(0)
            ans += 1
            for dx, dy in zip(dxs, dys):
                nx, ny = x + dx, y + dy
                if 0 <= nx < len(grid) and 0 <= ny < len(grid[0]):
                    if grid[nx][ny]:
                        grid[nx][ny] = 0
                        queue.append((nx, ny))
        return ans

 

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