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标题: 突然想到让deepseek来解释一下递归 [打印本页]

作者: 密银    时间: 2025-1-29 14:16
标题: 突然想到让deepseek来解释一下递归
本帖最后由 密银 于 2025-1-29 14:19 编辑
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解释的不错  J; A$ J8 {- A! h( ]: q! ~
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递归是一种通过将问题分解为更小的同类子问题来解决问题的方法。它的核心思想是:**函数直接或间接地调用自身**,直到满足终止条件。* I/ `6 p! }5 \! b/ o0 }% H* E

) ~1 W: j# a: f4 T+ X% s2 h. t0 q 关键要素. Z6 x/ L, j$ Y) l7 w
1. **基线条件(Base Case)**& x  p9 U* R8 ^7 k# b! S9 m
   - 递归终止的条件,防止无限循环8 F+ K3 \* f- o, \, M
   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1
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: @: G# t: W% W8 F- Y% f/ H2. **递归条件(Recursive Case)**
; P' m; J  q0 h6 T6 ~) ]& ]9 g; w   - 将原问题分解为更小的子问题/ Y# q: Y" W' ?7 }) W
   - 例如:n! = n × (n-1)!
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; A1 z) ?5 i6 H9 F6 w, q6 Z9 I 经典示例:计算阶乘
0 h# I. |3 u! r8 x! w- kpython
8 U0 ~, \& p) @" I  ^! Bdef factorial(n):
% w' k1 _+ {* a( u4 t/ [$ X# ]& K    if n == 0:        # 基线条件
( q! {3 Z/ Y7 U        return 1; o; }6 s  v' t8 b
    else:             # 递归条件. R- F6 Z8 L7 `0 `3 _  O
        return n * factorial(n-1)# G% A3 ^1 Q7 D, v3 B
执行过程(以计算 3! 为例):9 @; n/ O) \. S# u
factorial(3)
& J# T$ c$ w& I3 * factorial(2)* F, A! s$ C/ T, m/ n: d
3 * (2 * factorial(1))
) k  h6 R" t( t: h, a" F3 * (2 * (1 * factorial(0)))
  ?3 c; U- r! U3 * (2 * (1 * 1)) = 6! a5 b( a% A* f0 I5 h

" _( V4 e5 A  }# u- H: h2 v 递归思维要点
# a- M! w4 a3 {/ Y& z1 r0 G0 p$ G1. **信任递归**:假设子问题已经解决,专注当前层逻辑) r- y, U" ^  @  t0 g" M# k
2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
% G, o4 u8 n* ]4 G8 r8 K8 [3. **递推过程**:不断向下分解问题(递)
8 R$ t1 _/ D8 |, J: |2 Q4. **回溯过程**:组合子问题结果返回(归)3 S$ d4 @5 ~: `$ j# ^/ R
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注意事项
7 j6 P7 l9 l% D必须要有终止条件
1 e3 z! y' e3 O& H! k! d递归深度过大可能导致栈溢出(Python默认递归深度约1000层)2 b( i7 U8 V+ K
某些问题用递归更直观(如树遍历),但效率可能不如迭代! n4 b' n* K$ o0 K! g
尾递归优化可以提升效率(但Python不支持)
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递归 vs 迭代5 s% z* C1 K6 u  m" R7 C% [$ n
|          | 递归                          | 迭代               |
: f9 b- R5 u* F; Z|----------|-----------------------------|------------------|
3 e; ~/ t; [* y: X| 实现方式    | 函数自调用                        | 循环结构            |9 v( z$ Z5 Q4 v3 O; a) n
| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |% c4 v5 o+ L4 g" u
| 代码可读性  | 对符合递归思维的问题更直观                | 线性流程更直接         |. A0 |0 Y  X6 L+ Z/ F2 R6 T7 k
| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |
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. b  Q/ X8 v* `5 i! `7 x9 \ 经典递归应用场景
8 d) R- a) T$ x, }0 O+ R% [6 O, a2 r1. 文件系统遍历(目录树结构)/ L& K* L* l- G7 H& f5 s
2. 快速排序/归并排序算法; e; r* K. y+ h; D# O$ H
3. 汉诺塔问题3 w1 Z+ V( `! w4 l1 B! w
4. 二叉树遍历(前序/中序/后序)( J( q& S7 e7 A! V  @
5. 生成所有可能的组合(回溯算法)/ C; e( @" z* {3 Q7 y

$ S$ ^/ n7 Q( \4 k5 ~! }试着用递归思维想象:你站在一面镜子前,镜子里有无数个逐渐变小的你在照镜子,这就是递归的直观体现。但记住每个"分身"最终都要有结束的时刻,这就是基线条件的重要性。
作者: testjhy    时间: 2025-1-30 00:07
挺好,递归思维要点与我能够回忆起来我当时写递归程序的思路很一致,,或者被它唤醒," R/ E  |6 z1 f+ L2 A- o
我推理机的核心算法应该是二叉树遍历的变种。4 G8 D7 G  ~5 K2 q/ {/ {/ U
另外知识系统的推理机搜索深度(递归深度)并不长,没有超过10层的,如果输入变量多的话,搜索宽度很大,但对那时的286-386DOS系统,计算压力也不算大。
作者: nanimarcus    时间: 2025-2-2 00:45
Recursion in programming is a technique where a function calls itself in order to solve a problem. It is a powerful concept that allows you to break down complex problems into smaller, more manageable subproblems. Here's a detailed explanation:
: Y, E1 p3 A& X2 Z7 b* rKey Idea of Recursion
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) D5 `3 ]- v( M- F  |) m; wA recursive function solves a problem by:
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. d" y5 ~/ @3 R6 y5 V5 g' \) K    Breaking the problem into smaller instances of the same problem.
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% v3 F- m8 o& [8 P    Solving the smallest instance directly (base case).
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    Combining the results of smaller instances to solve the larger problem.
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Components of a Recursive Function
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    Base Case:
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        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.$ K4 |: P9 ]( W: @! i: |
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        It acts as the stopping condition to prevent infinite recursion.2 i) B# D! f8 z+ M
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        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.
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    Recursive Case:- T4 e& z. }# `; D
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        This is where the function calls itself with a smaller or simpler version of the problem.
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        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).
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Example: Factorial Calculation* ?' D1 I) E5 Y9 d" o  P. I2 a

" w' g) J; z4 m2 XThe factorial of a number n (denoted as n!) is the product of all positive integers less than or equal to n. It can be defined recursively as:, c5 c- v8 k8 ?* v  |# |" g' a
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    Base case: 0! = 1
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1 k/ m" ^6 q. o    Recursive case: n! = n * (n-1)!9 h2 M# w- z5 H6 M/ d8 s- R9 C' z
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Here’s how it looks in code (Python):
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( \/ B& i( S, i; o- cdef factorial(n):: i2 c4 k0 P) r
    # Base case* `1 z# t! I& [2 a
    if n == 0:/ j" O( ^0 P, }; g* ]
        return 1
7 J7 {* X& R: i# p  ]9 E- i2 \8 c    # Recursive case
, A* c7 r' N0 \6 y4 V9 ^. s4 c    else:
, q+ [3 o) O! J8 C" d        return n * factorial(n - 1)
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8 r4 t, {8 o, F9 `; G1 b# Example usage% L$ {0 Q5 t7 W
print(factorial(5))  # Output: 120
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How Recursion Works
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    The function keeps calling itself with smaller inputs until it reaches the base case.
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    Once the base case is reached, the function starts returning values back up the call stack.8 I  w; S- D9 N6 ^5 D  Z4 U
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    These returned values are combined to produce the final result.0 A& U6 J, ~$ q6 ^/ i& d# a' a

* u& q$ T' v! o0 U4 E  KFor factorial(5):- S6 P$ H! ~% _

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+ ?, S. O9 C- \1 Gfactorial(5) = 5 * factorial(4)$ m- q5 r7 Y* B) W6 t$ ^* ~; Z
factorial(4) = 4 * factorial(3)8 K- E# n" p8 M" F1 t8 o
factorial(3) = 3 * factorial(2)
! [; N' E! x3 k+ [% J: ufactorial(2) = 2 * factorial(1)  T5 `+ j4 ]% j" s1 t" v: f
factorial(1) = 1 * factorial(0)' V- S5 `# U5 g" M
factorial(0) = 1  # Base case0 [) j4 |, m; q" [! t# ?9 P

$ E2 n! o4 k( S$ MThen, the results are combined:
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# P8 S5 B) O, ~& efactorial(1) = 1 * 1 = 11 V4 x# e. D2 Y
factorial(2) = 2 * 1 = 2" W6 L. o: _+ [' F$ Y0 @1 r2 Y
factorial(3) = 3 * 2 = 6
' j! i# r  g7 ^. [. T9 S' m* z, ?0 dfactorial(4) = 4 * 6 = 244 \* h$ b$ I$ Z; u& A
factorial(5) = 5 * 24 = 120
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Advantages of Recursion# h+ }6 d# H& _# c3 A  {$ `
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    Simplicity: Recursive solutions are often more intuitive and easier to write for problems that have a natural recursive structure (e.g., tree traversals, divide-and-conquer algorithms).5 L$ f( C# n2 b& P, R4 H3 l

. t4 n6 b. A- v    Readability: Recursive code can be more readable and concise compared to iterative solutions.: e! ]$ L) ?0 Q2 a4 ^" `$ |  `

% ~# \7 A/ N) n0 WDisadvantages of Recursion; M7 l0 Y, F% y: X& U( @% c
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    Performance Overhead: Each recursive call adds a new layer to the call stack, which can lead to high memory usage and potential stack overflow for deep recursion.7 f( c4 k8 Z! w8 q/ C; l0 T

! R! k1 D2 p7 O: f/ a2 P. j  K1 k    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
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When to Use Recursion
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    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
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# U0 `& y" I$ s8 l2 l    Problems with a clear base case and recursive case.' o# |; K5 a+ n, x+ J

, E# Y' P& K  x9 [Example: Fibonacci Sequence
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) i; Y; b2 a+ h3 w9 [The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
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( d8 |6 f6 [' r$ [, ~7 T6 n    Base case: fib(0) = 0, fib(1) = 1
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    Recursive case: fib(n) = fib(n-1) + fib(n-2)9 E9 _3 ~- }7 v% }

' Q% c* F5 d6 V' U6 A, E% xpython
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; g0 N: {6 T% `  I, m4 n: kdef fibonacci(n):
% @8 c- T+ o5 y2 I% ?7 ?    # Base cases, P$ v) N. v/ w6 y# B
    if n == 0:+ l" D/ d9 o) }, I
        return 0! ?" ?9 k  N( b7 S4 Y
    elif n == 1:
" y. J! c. j8 I! B6 j* s        return 1
& E8 t' |/ U% A! ]' Z    # Recursive case4 |/ @3 f" z, }2 K# \/ w
    else:
0 e; O+ N% C. C; }: g        return fibonacci(n - 1) + fibonacci(n - 2)+ @" b4 ?: a- n2 K0 O
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# Example usage
3 H  l# e/ k. _0 Qprint(fibonacci(6))  # Output: 8( a' F3 q, c& U3 u
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Tail Recursion
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( \0 {, Y' G7 |' C1 L  m. B$ oTail recursion is a special case of recursion where the recursive call is the last operation in the function. Some programming languages optimize tail-recursive functions to avoid stack overflow, but not all languages (e.g., Python does not optimize tail recursion).
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In summary, recursion is a fundamental concept in programming that allows you to solve problems by breaking them into smaller, self-similar subproblems. It’s important to define a base case to avoid infinite recursion and to understand the trade-offs between recursion and iteration.
作者: nanimarcus    时间: 2025-2-2 00:47
我还让Deepseek 给我讲讲Linux Kernel Driver 现在的开发流程,让一个老同志复习复习,快忘光了。




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