深度优先搜索 (DFS) 和广度优先搜索 (BFS)是比较难的算法问题，但也是面试常考题，因此需要认真研究并掌握。

DFS 用递归实现，BFS用栈实现

1. 岛屿问题（岛屿连通）

1.1 岛屿数量

LeetCode链接：LeetCode 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

1.1.1 DFS 解法

code：

class Solution:def numIslands(self, grid: List[List[str]]) -> int:# dfs 用递归实现def dfs(i,j):if not 0<=i<len(grid) or not 0<=j<len(grid[0]) or grid[i][j]!="1":return# 遍历完(i,j)后将其标记为水，防止重复搜索grid[i][j]="0"vectors=([i,j+1],[i-1,j],[i,j-1],[i+1,j])for vector in vectors:dfs(vector[0],vector[1])num=0for i in range(len(grid)):for j in range(len(grid[0])):if grid[i][j]=="1":dfs(i,j)num+=1return num

1.1.2 BFS 解法

将 BFS 写成函数的代码写法：

class Solution:def numIslands(self, grid: List[List[str]]) -> int:# 通过bfs将搜索过的节点标记为“0”# bfs通过栈实现（先进先出）def bfs(i,j):if not 0<=i<len(grid) or not 0<=j<len(grid[0]) or grid[i][j]!="1":returnstack.append([i,j])grid[i][j]="0"while stack:stack.pop(0)vectors=([i,j+1],[i-1,j],[i,j-1],[i+1,j])for vector in vectors:bfs(vector[0],vector[1])num=0for i in range(len(grid)):for j in range(len(grid[0])):if grid[i][j]=="1":stack=[]bfs(i,j)num+=1return num

使用迭代而非函数的 BFS 写法：

class Solution:def numIslands(self, grid: List[List[str]]) -> int:num=0for i in range(len(grid)):for j in range(len(grid[0])):if grid[i][j]=="1":# bfs通过栈实现（先进先出）# deque 是双向队列，可以高效的在队列头部和尾部添加、删除元素deque=collections.deque([[i,j]])grid[i][j]="0"     # 将搜索过的节点标记为“0”while deque:row,col=deque.popleft()vectors=([row,col+1],[row-1,col],[row,col-1],[row+1,col])for vector in vectors:if 0<=vector[0]<len(grid) and 0<=vector[1]<len(grid[0]) and grid[vector[0]][vector[1]]=="1":deque.append(vector)grid[vector[0]][vector[1]]="0"    # 将搜索过的节点标记为“0”num+=1return num

注：

在使用先入先出栈时，应优先使用 collections.deque() 而非 list