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)