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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 ( u/ [1 Z! H: i9 }" }5 O: n

. D5 c$ v% s+ u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。* v' ]0 `8 t) O+ Q$ l! k; I: i

0 ^! f& s! w9 N+ }. M& d/ M 关键要素. q/ ~/ `3 H+ }4 f
1. **基线条件(Base Case)**
2 T" i3 M- F2 H8 T: F. s' j   - 递归终止的条件,防止无限循环* S5 @# q. \( f
   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1/ o3 `0 ?" H: A
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2. **递归条件(Recursive Case)**
/ D# }' h1 b* }0 z; d   - 将原问题分解为更小的子问题' ~  _- ]- a! X6 i
   - 例如:n! = n × (n-1)!6 Q  |7 p7 e) {$ Q2 B0 l

0 Y# u, P, S3 T/ h7 c8 d 经典示例:计算阶乘
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def factorial(n):; C  j1 [! \: `8 K5 z6 d
    if n == 0:        # 基线条件
! F  j5 p4 A( s; |+ [/ n        return 1' }1 e6 R% P( H& l7 q
    else:             # 递归条件  u# |. R3 B1 c9 z; N8 k* O
        return n * factorial(n-1)
, K6 z' B; d( _执行过程(以计算 3! 为例):
' }2 _# \7 x5 q  U. w. y& d( pfactorial(3)
4 K3 @8 e4 A" H1 e3 * factorial(2)
9 e' W4 E# v& Z) v4 X2 `/ E3 * (2 * factorial(1))
* t) L! k% j8 Q$ m3 * (2 * (1 * factorial(0)))1 V8 n! G  [$ Q7 R- N
3 * (2 * (1 * 1)) = 6
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递归思维要点
0 f/ U! \$ U1 {: a/ i# B* x1. **信任递归**:假设子问题已经解决,专注当前层逻辑
' G- x7 R  v5 J" f# g2. **栈结构**:每次调用都会创建新的栈帧(内存空间), U8 @0 w2 _9 \
3. **递推过程**:不断向下分解问题(递)
& i& g3 ~5 q- O9 D4. **回溯过程**:组合子问题结果返回(归)
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注意事项7 a% U# @! c- O8 i
必须要有终止条件
0 A7 }- z/ s0 F7 B8 E递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
4 @$ d1 x. J4 i: E( x某些问题用递归更直观(如树遍历),但效率可能不如迭代& X5 Z, J  o2 H9 ^2 C
尾递归优化可以提升效率(但Python不支持)+ P  Y+ W$ |# ]" Z' {" l( L

% c- j3 Z  C: d. v 递归 vs 迭代
( R2 c# y% a0 D$ a, C1 u6 S/ l|          | 递归                          | 迭代               |
* {* H: e( \1 Z# {9 T|----------|-----------------------------|------------------|
/ |' ]& l* y$ q7 T| 实现方式    | 函数自调用                        | 循环结构            |
0 t) O5 f7 t3 n& t| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |* c/ C; |! g. X( E& M5 A
| 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         |
) x, ]7 n$ n/ ^9 L; G| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |  J) C' v% _8 C( S/ s$ F! C# L
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经典递归应用场景
) v; r) x: |7 p1. 文件系统遍历(目录树结构)( W2 e  x, B0 _1 s+ z
2. 快速排序/归并排序算法2 J( {: V. p: _1 C2 Z
3. 汉诺塔问题
* `1 [- g* S! O4. 二叉树遍历(前序/中序/后序)  p! ^8 g# X1 F. Q7 Z$ `3 z$ G0 F
5. 生成所有可能的组合(回溯算法)
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6 W, e3 y9 N3 y1 {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,
* F1 F7 Q. p& R( c1 R我推理机的核心算法应该是二叉树遍历的变种。
* Z/ l1 `" C4 }; A4 m4 b! ~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:' e/ m; T" a7 R& v9 h+ a4 S$ _
Key Idea of Recursion1 b7 v. D1 M" I

/ M" f# k9 D" }: N  ~& S  r, y# v0 iA recursive function solves a problem by:- o. Z3 r" }  H8 H' o! V

9 e' l2 w$ X& u, q( R    Breaking the problem into smaller instances of the same problem.
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    Solving the smallest instance directly (base case).
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( ^, R: r5 T* C- D    Combining the results of smaller instances to solve the larger problem.
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( g! G# L5 `9 s$ {6 S. ~7 NComponents of a Recursive Function1 E. Q3 q' r, X3 x
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    Base Case:8 K7 {& U" ]3 I. p9 |. Q

& t8 b/ U1 E& n& O# X7 T2 e* Z        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.
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        It acts as the stopping condition to prevent infinite recursion.. f; |( X  O( q% P, ^5 @' u( C
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        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.$ ?9 N8 r1 L3 F8 M
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    Recursive Case:
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/ W% U+ L; f1 T; |$ B# j        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 Calculation7 Y# R" f8 v7 N  ]* w. @8 J' p
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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:, j' Z. I1 N  ]8 h. l

3 q  Q; ^. s. X0 I1 W. G' T    Base case: 0! = 1
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$ I, L- g, s3 t! n' a! @, y    Recursive case: n! = n * (n-1)!7 B$ \- w  y! j7 L! H7 X$ K

( A" i" ~3 ?3 a* ^) SHere’s how it looks in code (Python):% B  y* {- l, `0 n
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def factorial(n):
2 }( t# K: w) d8 X/ S, p2 T4 _1 D$ M    # Base case
2 [. D5 h6 ]1 i2 T' M! U    if n == 0:
+ I* p; N; M5 r        return 1, _4 @  J, w  I: R
    # Recursive case3 R# l0 O7 t9 w3 \0 p) {
    else:
% m, p$ T6 H0 t8 G6 x+ A7 M        return n * factorial(n - 1)
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# Example usage. H5 }$ U3 T( Y" I" Q
print(factorial(5))  # Output: 120  y/ p- v# S- u+ A
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How Recursion Works: L1 x( r) f! J" E

" v' k- G8 h& D* E( B) c8 r    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.4 l- |7 b8 s' U' k/ X: h: r

0 t. L! B0 w, O& m+ J" p$ `) E  W3 [    These returned values are combined to produce the final result.
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For factorial(5):
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factorial(5) = 5 * factorial(4)3 a% L1 U. S* x3 W& j
factorial(4) = 4 * factorial(3)
8 ^1 W- Q  L6 q+ Cfactorial(3) = 3 * factorial(2)+ b5 H0 j! m0 B. C3 F8 s3 n1 W8 f
factorial(2) = 2 * factorial(1)
+ h4 @0 `1 F$ \factorial(1) = 1 * factorial(0); O) O) m! P7 ]: H
factorial(0) = 1  # Base case9 Q! Q8 i" Y' h* P

# I6 o' @, N/ f2 H" g7 P6 LThen, the results are combined:9 R9 E! b8 _- H

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factorial(1) = 1 * 1 = 1
% A& q5 l9 w( ifactorial(2) = 2 * 1 = 2
" e1 P3 e* F+ i; h5 O: ffactorial(3) = 3 * 2 = 6' R& }1 C' h* l; j9 f2 ]
factorial(4) = 4 * 6 = 24" U" u0 k3 l4 o$ g) m& c# B: `
factorial(5) = 5 * 24 = 120
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Advantages of Recursion5 |. P0 K5 t: [3 q! |. ~

- o; b; [" E1 \8 N3 @) x! i    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).* t- m' U; K9 P# [3 N7 i8 i

! `% n8 @+ b, z/ z$ u9 U9 A    Readability: Recursive code can be more readable and concise compared to iterative solutions.
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7 S# K3 l( O' ]/ `; k0 t* K) ?# _7 ADisadvantages of Recursion
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) J+ W2 u# H" ?; P( n    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.
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    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).9 I5 B" x. I3 v/ n5 Z# _
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When to Use Recursion
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5 I' @4 M1 t# F! Y* O/ p7 ?    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
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4 N: o- n4 G- T, K    Problems with a clear base case and recursive case.5 J4 w% o( _: h6 a3 y* t5 Y

- S9 r: M; a. I1 L) P; yExample: Fibonacci Sequence0 l( d3 Z( E4 Z1 }: u8 f( o% W
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The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:$ d6 V- {4 E% j( l$ Z4 F
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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)2 B% Z' T8 ^& F2 i! q* G

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def fibonacci(n):4 m* a$ |1 T! O1 B4 \3 L4 _
    # Base cases. O( v/ Y" Q+ I0 z! J/ a
    if n == 0:
8 R/ Y! @- o$ u3 ?. N( s* `        return 0& ]" H4 q- L# r  K. O- {, Y
    elif n == 1:( |, }) G: |8 y) E
        return 1
- Q9 s% ^& z1 _+ B. S" [    # Recursive case
9 p! G6 y2 }) q    else:# x2 s& e: Q& q! C
        return fibonacci(n - 1) + fibonacci(n - 2)3 I3 Y4 N+ M6 }# p2 W- p3 c: a+ n
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# Example usage
, G0 u! f7 x% N4 n1 t; Zprint(fibonacci(6))  # Output: 8
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8 M7 b, K) b6 I' XTail Recursion. b# @$ B- E. S, V! [! z& d

5 `2 V+ v0 w- T2 X" U2 h( ]5 TTail 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).& O; X9 v+ F' f8 m$ \

/ ~% M* i9 ]9 J3 e* L: iIn 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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