【宽度优先搜索 BFS】Leecode-200. 岛屿数量

【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

【题解】

题解1：

• 思路

     列出可能的树形图，对于给出的列表逐个遍历调用dfs

• 复杂度

         时间复杂度：O(n2)，空间复杂度：O(n)

• 代码

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        if not grid:
            return []
        m = len(grid)
        n = len(grid[0])
        def bfs(grid, i, j):
            queue = [[i,j]]
            while queue:
                [i,j] = queue.pop(0)
                if 0<=i<m and 0<=j<n and grid[i][j]=='1':
                    grid[i][j] = '0'
                    queue.extend([[i+1, j], [i-1,j], [i,j+1], [i, j-1]])
        count=0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '0':
                    continue
                bfs(grid, i, j)
                count += 1
        return count