题目
给定一个包含了一些 0 和 1的非空二维数组 grid , 一个 岛屿 是由四个方向 (水平或垂直) 的 1 (代表土地) 构成的组合。你可以假设二维矩阵的四个边缘都被水包围着。
找到给定的二维数组中最大的岛屿面积。(如果没有岛屿,则返回面积为0。)
示例 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]]
对于上面这个给定矩阵应返回 6。注意答案不应该是11,因为岛屿只能包含水平或垂直的四个方向的‘1’。
示例 2:
[[0,0,0,0,0,0,0,0]]
对于上面这个给定的矩阵, 返回 0。
注意: 给定的矩阵grid 的长度和宽度都不超过 50。
解答
典型的搜索题,深度优先和广度优先都可以。(比较简单,不记录思路了)
1,深度优先dfs,用递归。Time: O(mn),Space:O(mn)
2,广度优先bfs,用队列。Time: O(mn),Space:O(mn)
# 深搜
class Solution:
def __init__(self):
self.a = []
self.book = []
self.next = [
[0, 1],
[1, 0],
[0, -1],
[-1, 0]
]
self.m = 0
self.n = 0
self.max = 0
def maxAreaOfIsland(self, grid) -> int:
if not grid:
return 0
self.a, self.m, self.n = grid, len(grid), len(grid[0])
self.book = [
[0 for _ in range(self.n)] for _ in range(self.m)
]
max = 0
for i in range(self.m):
for j in range(self.n):
if self.a[i][j] == 1:
self.max = 0
self.book[i][j] = 1
self.max += 1
self.dfs(i, j)
if self.max > max:
max = self.max
return max
def dfs(self, x, y):
for i in range(4):
tx = x + self.next[i][0]
ty = y + self.next[i][1]
if tx >= 0 and ty >= 0 and tx < self.m and ty < self.n and self.book[tx][ty] == 0 and self.a[tx][ty] == 1:
self.max += 1
self.book[tx][ty] = 1
self.dfs(tx, ty)
# 不取消标记,只走一次即可
s = Solution()
ans = s.maxAreaOfIsland([[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]])
print(ans)
# 6