刷题记录 HOT100 图论-1：200. 岛屿数量

题目：200. 岛屿数量

难度：中等

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'

一、模式识别

1.图论

岛屿问题 》 图论经典问题 》

方法：遍历数组，找到陆地后搜索出整片陆地，防止在以后的搜索中只找新大陆

陆地处理方法： （访问）数组法、缩地法

数组法：用visited数组标记访问过的陆地，每遇见新大陆就把整片大陆标记已访问

缩地法：原地修改grid数组，每遇见新大陆就把整片新大陆填平

搜索方法：DFS和BFS均可

二、代码实现

1.数组法

（1）DFS

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        def dfs(i, j):
            for di, dj in dirs:
                ti, tj = i + di, j + dj 
                if (0 <= ti < m and 0 <= tj < n) and grid[i][j] == "1" and not visited[ti][tj]:
                    visited[ti][tj] = True
                    dfs(ti, tj)

        dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        m, n = len(grid), len(grid[0])
        visited = [[False] * n for _ in range(m)]

        ans = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == "1" and not visited[i][j]:
                    ans += 1
                    visited[i][j] = True
                    dfs(i, j)
        return ans

（2）BFS

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        def bfs(i, j):
            que = collections.deque([(i, j)])
            while que:
                ni, nj = que.popleft()
                for di, dj in dirs:
                    ti, tj = ni + di, nj + dj
                    if (0 <= ti < m and 0 <= tj < n) and grid[ti][tj] == "1" and not visited[ti][tj]:
                        visited[ti][tj] = True
                        que.append((ti, tj))

        m, n = len(grid), len(grid[0])
        dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        visited = [[False] * n for _ in range(m)]
        ans = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == "1" and not visited[i][j]:
                    ans += 1
                    visited[i][j] = True
                    bfs(i, j)
        return ans

2.缩地法

 （1）DFS

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        def dfs(i, j):
            for di, dj in dirs:
                ti, tj = i + di, j + dj
                if (0 <= ti < m and 0 <= tj < n) and grid[ti][tj] == "1":
                    grid[ti][tj] = "0"
                    dfs(ti, tj)

        m, n = len(grid), len(grid[0])
        dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        ans = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == "1":
                    ans += 1
                    grid[i][j] = "0"
                    dfs(i, j)
        return ans

（2）BFS

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        def bfs(i, j):
            que = collections.deque([(i, j)])
            while que:
                ni, nj = que.popleft()
                for di, dj in dirs:
                    ti, tj = ni + di, nj + dj
                    if (0 <= ti < m and 0 <= tj < n) and grid[ti][tj] == "1":
                        grid[ti][tj] = "0"
                        que.append((ti, tj))

        m, n = len(grid), len(grid[0])
        dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        ans = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == "1":
                    ans += 1
                    grid[i][j] = "0"
                    bfs(i, j)
        return ans