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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 5 F% i! E' t( M$ t
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解释的不错
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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。: N  Y! n# b. y3 I& H) |
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关键要素% u: q* L" `" B  E, z1 j
1. **基线条件(Base Case)**- r* Z1 q, |( O  s- N0 K/ v; ?% p0 r
   - 递归终止的条件,防止无限循环3 K: c+ U$ E6 I2 @! Y1 H
   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1
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) P6 ]2 Y( F& i) U( X6 _2. **递归条件(Recursive Case)**
4 }0 u; C' r- ]2 K8 m6 t   - 将原问题分解为更小的子问题2 `% f6 ~' X) R
   - 例如:n! = n × (n-1)!2 Y, v. m: Q7 X: a4 C& h  Z5 M
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经典示例:计算阶乘
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6 }' J$ K) \7 H( p: U: Odef factorial(n):1 p) w+ D' _# `9 U& u5 Q4 k2 E
    if n == 0:        # 基线条件  R9 w' j  G  D5 P4 ~( p
        return 1
, o9 Q$ \. _; r9 d    else:             # 递归条件
4 k# E; m3 {1 ?& h        return n * factorial(n-1)9 u2 ^( M( _+ N) E7 k
执行过程(以计算 3! 为例):
+ f( f3 E0 t& p! V4 |9 efactorial(3). c) D0 ?2 ~( Z5 W6 s
3 * factorial(2)
* \/ q' q2 w; t8 e3 * (2 * factorial(1))' _; q/ i3 k& o& K# t$ g' N' R  u( O
3 * (2 * (1 * factorial(0)))
1 e$ X( |4 t5 o% B1 v8 ~+ z- C3 * (2 * (1 * 1)) = 6
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; v) U& l( O* Q* k, Q 递归思维要点
( ~  C- n, b6 P1 u/ z" e1 z; z1. **信任递归**:假设子问题已经解决,专注当前层逻辑
; y+ z) G9 o( o! ]0 `2. **栈结构**:每次调用都会创建新的栈帧(内存空间)8 G3 X7 z1 t+ L
3. **递推过程**:不断向下分解问题(递)
; s, k2 p+ d8 C4 C4. **回溯过程**:组合子问题结果返回(归)/ p' y* i. ~  |6 H3 y

3 w. V- h5 v9 Y# p: x 注意事项2 X. K3 V' H# Q2 i: u
必须要有终止条件3 ~' z  p# e5 {& H4 _6 @
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
+ N* ^3 A, r( J4 `" D7 m* Q) v某些问题用递归更直观(如树遍历),但效率可能不如迭代
- J4 n$ f* m2 d' g) G尾递归优化可以提升效率(但Python不支持)8 \- B1 X. _) A5 E. p
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递归 vs 迭代6 `, }9 A6 X/ ~" c
|          | 递归                          | 迭代               |1 Z, J& {0 C4 L& W3 A
|----------|-----------------------------|------------------|; u1 G* @! |1 m+ x! V: h
| 实现方式    | 函数自调用                        | 循环结构            |
$ N0 ~- }6 \( \6 r) R| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
' ^5 n# K7 q, Y3 K5 i| 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         |& u$ t" h! n+ S% `# `& P, n
| 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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经典递归应用场景2 J3 c/ s# u8 j* [' J
1. 文件系统遍历(目录树结构)3 c+ K* C' |! d) Y
2. 快速排序/归并排序算法
3 t1 M0 @3 E6 s$ |4 u3. 汉诺塔问题
  Q6 m" ^: z9 E1 ]8 x0 ]( A4. 二叉树遍历(前序/中序/后序)
5 j3 |5 |1 V4 q# f& d5. 生成所有可能的组合(回溯算法)
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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,
3 B. k  T" d$ ^& h* E7 q- }, X我推理机的核心算法应该是二叉树遍历的变种。
- F' S5 t1 P" B- j4 [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:
* l5 {, ?  @2 }+ s: i: m2 }Key Idea of Recursion2 d4 k+ l( K) I; e8 z

# t5 f9 |# D5 p! ?% hA recursive function solves a problem by:( _( f9 P( g' Q4 g/ y8 M

: X' ?3 i9 T4 w6 R    Breaking the problem into smaller instances of the same problem.  N3 C! ^! g: M* G  \- W. R6 H
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    Solving the smallest instance directly (base case).' L1 ~" T) _; c( D/ v- h
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    Combining the results of smaller instances to solve the larger problem.
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1 Q2 d- `( R$ Y5 |5 w7 q; lComponents of a Recursive Function
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5 v! I/ p  V9 ^# a8 M- V. k    Base Case:
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        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.) L$ P  T  G. m) D7 k

9 L0 [8 m7 z5 M; V2 {        It acts as the stopping condition to prevent infinite recursion.
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        Example: In calculating the factorial of a number, the base case is factorial(0) = 1." q6 n7 A2 `* a
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    Recursive Case:6 E% ]% M, |9 m: m

4 g# N" R- i+ |) R; }/ }) n2 g' J1 i        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).- G, L* W5 d+ ~: S) P/ v9 T
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Example: Factorial Calculation
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The 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:) X% f) M8 v6 W- g1 _* ]
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    Base case: 0! = 11 _8 Y7 e3 t# G8 ^, i* w6 @
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    Recursive case: n! = n * (n-1)!
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Here’s how it looks in code (Python):  ~" C$ p) K% z
python
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0 \- G2 A2 _( G3 W) K+ x; Ldef factorial(n):
0 p8 {) a; P& B- c) f/ m! M    # Base case$ L  E! f, V* ^9 \& U! M8 \# U% j. b( I
    if n == 0:
3 E/ @" p) G2 w" @6 e% s        return 1
( f: O5 B) S/ z. ?    # Recursive case
1 i. n: w; ]0 U' T- v# X+ f    else:
; O- Q( Q, _5 \        return n * factorial(n - 1)
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7 b$ r1 F( v9 z/ r) Y% e( m# Example usage
0 U) X5 n- j0 S2 t0 ?- R* C+ q0 O! Lprint(factorial(5))  # Output: 120) N; `( R/ K# t' {

8 c" ]- n% z. P& W" W0 LHow Recursion Works
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    The function keeps calling itself with smaller inputs until it reaches the base case.
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2 ?6 i- i, O% x3 [+ @8 |    Once the base case is reached, the function starts returning values back up the call stack.9 U  s# J1 q$ t+ y% A

8 d# Y) V  W$ E3 |    These returned values are combined to produce the final result.! N5 D1 i7 Y  ?# v% x# A# C1 D

" A- P. {" r" t% t$ j6 aFor factorial(5):4 k3 ~) h% s' [- A  e

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factorial(5) = 5 * factorial(4)
! P! Z4 A7 |# k  Q. {0 N9 o; P  _factorial(4) = 4 * factorial(3)
/ R8 Y0 J+ K- u3 b+ O! g6 C  ]factorial(3) = 3 * factorial(2)
2 ]2 B  ~. j' N( Qfactorial(2) = 2 * factorial(1)
* A' p0 ^! {2 |' C3 _5 ]0 M# lfactorial(1) = 1 * factorial(0)8 {' a# ?* K+ }+ G0 C
factorial(0) = 1  # Base case0 k% g4 e1 {% _' `. n1 Y

- v/ O& P0 K# m+ sThen, the results are combined:* V3 n3 \! S& ~1 V' K  Y" L4 x

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& `  A: ~& J$ h7 Z) O" u) v$ T) f' y: ]factorial(1) = 1 * 1 = 1+ }: D( G" Z$ Z( z
factorial(2) = 2 * 1 = 2/ `' z/ o8 _7 y- ^1 _& K
factorial(3) = 3 * 2 = 68 }/ P6 O2 o- s
factorial(4) = 4 * 6 = 242 `" e4 o- @9 X& W% p# \# E
factorial(5) = 5 * 24 = 120
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  Y" B: O' i9 QAdvantages of Recursion
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/ L6 R. B  t6 ?: x    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).+ K0 a8 I" A6 m4 L8 R# W

2 h/ \* X* ~# f, Q& S. ^' P    Readability: Recursive code can be more readable and concise compared to iterative solutions.
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- e7 V! |2 F0 o5 u( ?& KDisadvantages of Recursion* K' Q% V& s" d' ^9 G* c+ N1 f# {8 ~
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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." @# l4 g' V& V' h! S$ S
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    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).  t6 `+ `$ `" B/ a0 Z5 a

: v: D. E+ |- H$ gWhen to Use Recursion" Z+ B# [6 V/ k+ G- x, L
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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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, i8 i. r" Q3 Z# z. r2 d$ z    Problems with a clear base case and recursive case.
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& a7 ?1 k. H/ AExample: Fibonacci Sequence
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0 o* A0 `- `; ZThe Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:( R8 [$ f* O5 m9 d) a  o9 E; }' {, s
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    Base case: fib(0) = 0, fib(1) = 1
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    Recursive case: fib(n) = fib(n-1) + fib(n-2)+ S0 L  N1 r& c# A8 K& x: Q: F& b# D

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def fibonacci(n):
% S$ K1 W2 @9 y0 A& R: L6 D    # Base cases
# [8 T5 i! P/ r' v4 Y. c, H% {    if n == 0:
% F3 q- l5 p" x$ m# P% x: n' B        return 0- t0 |* F# n/ [: K
    elif n == 1:
8 y0 {2 c3 W. w' }* o4 G        return 1% [# V0 `( w1 c: x( R
    # Recursive case4 G8 k1 w/ i0 P1 c% [
    else:
, ~9 l  P; L  r/ t) k" a: D: L        return fibonacci(n - 1) + fibonacci(n - 2)
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# Example usage
5 y  L( w* S1 X: }" i2 Jprint(fibonacci(6))  # Output: 8' d; c  h. D) p6 Z9 i
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Tail Recursion
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Tail 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).  K9 j, m" c( ?1 v
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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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