1. 题目
给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
此外，你可以假设该网格的四条边均被水包围。
示例 1：
输入：grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
输出：1
示例 2：
输入：grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
输出：3
提示：
m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'
2. 题解
2.1 解法1：DFS递归方法
from typing import List
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if not grid or not grid[0]:
return 0
m = len(grid)
n = len(grid[0])
ans = 0
for i in range(m):
for j in range(n):
if grid[i][j] == "1":
ans += 1
self.dfs(grid, m, n, i, j)
return ans
def dfs(self, grid, m, n, r, c):
grid[r][c] = "0"
if r - 1 >= 0 and grid[r - 1][c] == "1":
self.dfs(grid, m, n, r - 1, c)
if r + 1 < m and grid[r + 1][c] == "1":
self.dfs(grid, m, n, r + 1, c)
if c - 1 >= 0 and grid[r][c - 1] == "1":
self.dfs(grid, m, n, r, c - 1)
if c + 1 < n and grid[r][c + 1] == "1":
self.dfs(grid, m, n, r, c + 1)
2.2 解法2：DFS迭代方法
from typing import List
# DFS 迭代
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if not grid or not grid[0]:
return 0
m = len(grid)
n = len(grid[0])
ans = 0
for i in range(m):
for j in range(n):
if grid[i][j] == "1":
print(i, j)
ans += 1
self.dfs(grid, m, n, i, j)
return ans
def dfs(self, grid, m, n, row, col):
grid[row][col] = "0"
stack = [[row, col]]
while stack:
r, c = stack.pop()
if r - 1 >= 0 and grid[r - 1][c] == "1":
stack.append((r - 1, c))
grid[r - 1][c] = "0"
if r + 1 < m and grid[r + 1][c] == "1":
stack.append((r + 1, c))
grid[r + 1][c] = "0"
if c - 1 >= 0 and grid[r][c - 1] == "1":
stack.append((r, c - 1))
grid[r][c - 1] = "0"
if c + 1 < n and grid[r][c + 1] == "1":
stack.append((r, c + 1))
grid[r][c + 1] = "0"
if __name__ == "__main__":
s = Solution()
grid = [["1","1","1"],["0","1","0"],["0","1","0"]]
a = s.numIslands(grid)
print(a)
2.3 解法3：BFS迭代方法
from typing import List
from collections import deque
# BFS 迭代
class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if not grid or not grid[0]:
return 0
m = len(grid)
n = len(grid[0])
ans = 0
queue = deque()
for i in range(m):
for j in range(n):
if grid[i][j] == "1":
ans += 1
queue.append((i, j))
while len(queue) > 0:
top = queue.popleft()
r, c = top[0], top[1]
if r - 1 >= 0 and grid[r - 1][c] == "1":
queue.append((r - 1, c))
grid[r - 1][c] = "0"
if r + 1 < m and grid[r + 1][c] == "1":
queue.append((r + 1, c))
grid[r + 1][c] = "0"
if c - 1 >= 0 and grid[r][c - 1] == "1":
queue.append((r, c - 1))
grid[r][c - 1] = "0"
if c + 1 < n and grid[r][c + 1] == "1":
queue.append((r, c + 1))
grid[r][c + 1] = "0"
return ans
if __name__ == "__main__":
s = Solution()
grid = [["1","1","1"],["0","1","0"],["0","1","0"]]
a = s.numIslands(grid)
print(a)
