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[科技前沿] 突然想到让deepseek来解释一下递归

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  • TA的每日心情
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    2025-9-8 05:08
  • 签到天数: 3 天

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     楼主| 发表于 2025-1-29 14:16:55 | 只看该作者 回帖奖励 |倒序浏览 |阅读模式
    本帖最后由 密银 于 2025-1-29 14:19 编辑 3 h! o5 m3 e) X) @/ f

    : t9 X; Y6 |8 U3 d4 a% F解释的不错
    # h  i" q% Y& T) B0 u% [2 s1 ~" m& d* z1 J. q8 N
    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。- p1 o  Z; g* ~0 u' |. w5 ~
    ' }* s4 w. q0 v6 e
    关键要素
    4 E2 Q" d, K( {1 B1. **基线条件(Base Case)**
    : Q7 e) h: }0 L$ [2 D, _   - 递归终止的条件,防止无限循环
    # ^: T* ]  N) a8 A# l2 b   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1' s& [6 R3 ], \" v( b" U& ^- ~6 D

    8 L. c+ U3 k( |5 R2. **递归条件(Recursive Case)**6 i( b# e# x1 _6 V" W6 d4 O% A; O
       - 将原问题分解为更小的子问题
    * N( ?2 L6 e* Q- D: {   - 例如:n! = n × (n-1)!# |* G+ w7 z3 c0 S

    ) |) s6 o0 Z" N& z3 o# N; g& `3 Z 经典示例:计算阶乘
    4 Q) H8 w. j8 U+ b6 `3 b% Bpython
    " `1 Z' }2 B: W$ ]' zdef factorial(n):
    # l  `$ w2 Z- J& B$ g    if n == 0:        # 基线条件
    % P% w9 Z3 Q9 d/ g$ \4 A2 B        return 1+ B1 U/ r& m" i$ ^- E; Q7 a
        else:             # 递归条件
    6 L6 e6 A! G9 e  u; M9 ]        return n * factorial(n-1)4 r: q  P" W7 w# _* m) ~
    执行过程(以计算 3! 为例):
    1 z7 i6 M& W! X- o: ffactorial(3)+ h) r" W5 n% U/ e' S7 Q+ s
    3 * factorial(2)/ |* ~3 {2 K5 u' K- d4 N
    3 * (2 * factorial(1))
    ; n3 B5 |+ t4 o* |3 * (2 * (1 * factorial(0)))* d3 ?/ R) N6 U8 K8 W; U
    3 * (2 * (1 * 1)) = 6! [: h7 {3 F# Z: `, l0 M

    / Y# V/ z0 I6 d; O; F 递归思维要点$ a' y, O* B9 B2 v# v
    1. **信任递归**:假设子问题已经解决,专注当前层逻辑
    . {+ T1 z1 e" w2. **栈结构**:每次调用都会创建新的栈帧(内存空间)( {+ r( m% }: ^1 K9 D3 Q" s
    3. **递推过程**:不断向下分解问题(递)
    5 x  p; a. \* F1 v- K8 ?2 r$ Q; n4. **回溯过程**:组合子问题结果返回(归)9 y$ o. i, S0 f% f' b. X
    " I$ Q- G8 d5 u4 U9 [9 ~" g1 o
    注意事项; z6 ]) M( |& I8 ?5 E' X7 i
    必须要有终止条件$ g2 T! l4 s9 c! p
    递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
    7 Z+ T# g$ F) {) |7 f6 r; `. I' {某些问题用递归更直观(如树遍历),但效率可能不如迭代
    # |8 S" T2 K7 [3 k% u. @尾递归优化可以提升效率(但Python不支持)- J  c- I) y+ ?3 n; a7 m/ s+ F

    ( R2 Y  I* F' R 递归 vs 迭代3 i( w/ U% Y" c( a- b
    |          | 递归                          | 迭代               |
    ) K: z0 I% q/ i  M. E) Q|----------|-----------------------------|------------------|
    / k0 g' A: e0 h* H: N* C. V| 实现方式    | 函数自调用                        | 循环结构            |
    ; ?/ D2 {: i2 a* L$ W# d| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
    / o$ V( _9 K* d. b, y; L: v| 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         |
    # |2 ]3 k' S' u| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |
    & i2 i9 F" a6 s' w% a& P- a% W/ G7 z* ?6 ]+ _/ L! O
    经典递归应用场景+ H- j# Q. S' z6 m
    1. 文件系统遍历(目录树结构)
    ) U# g* @! q; v2. 快速排序/归并排序算法/ L! d/ V1 q! B  X2 |# r
    3. 汉诺塔问题3 N) o. I) h4 f) y9 h; c
    4. 二叉树遍历(前序/中序/后序)
    ' x" A4 ~8 F  Y9 x* h2 `5 Z5. 生成所有可能的组合(回溯算法)# n, c3 j# k0 x; B% ^. u

    4 d/ t! z, X( Z9 t+ |. H6 w/ W: Y+ C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。

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  • TA的每日心情

    昨天 09:23
  • 签到天数: 3174 天

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    沙发
    发表于 2025-1-30 00:07:50 | 只看该作者
    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,
    4 Z, [- u" s/ @+ L4 \# Y& C我推理机的核心算法应该是二叉树遍历的变种。) T# s1 g$ N: P9 N; T
    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。
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    板凳
    发表于 2025-2-2 00:45:59 | 只看该作者
    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:
    ' l0 }" s# _' _: i# CKey Idea of Recursion, O4 c' Z( o5 Y: K4 R

    # G1 D  q) x/ l* Q9 U# O% nA recursive function solves a problem by:
    0 G% X$ x. _4 L5 S( `7 O( X/ A4 K, K4 Y; q
        Breaking the problem into smaller instances of the same problem.% a; [7 R5 x4 g0 ?2 J; y( `

    ' O$ q4 j* y* z9 S    Solving the smallest instance directly (base case).
    $ }+ Z8 ?8 y( d: N$ K1 u6 W. b
    & d/ k) s, x2 y! x9 Q    Combining the results of smaller instances to solve the larger problem.
    * W# d3 G/ |7 B" ^
    , e: }$ Z2 [. z+ \! zComponents of a Recursive Function% I; U8 ^+ V3 q; d1 L/ b  j& n) b
    , E) S, C/ j$ d# Y' Z: J" {1 ]
        Base Case:
    + l7 r0 g: ~" X: ?6 u+ g$ f" n2 p0 P1 |! N8 a7 Z( W: r
            This is the simplest, smallest instance of the problem that can be solved directly without further recursion.
    $ R- g  q1 c" Q: a+ v! }7 K. e% Z6 E% C/ R0 z; P' T- Y5 X/ a
            It acts as the stopping condition to prevent infinite recursion.. v- ?  R  O$ q4 _. B0 @

    7 i" S/ n& {/ e- t        Example: In calculating the factorial of a number, the base case is factorial(0) = 1., Q/ b' W4 o8 P

    % w) w( [; u4 x    Recursive Case:3 j2 T( @( x7 ^0 t$ ?
    6 a0 i8 n; }2 q6 M* A! }
            This is where the function calls itself with a smaller or simpler version of the problem.
    3 c( i9 q4 N1 g4 g9 {2 W& X
    5 P% V9 `- A; u, ]/ T% Z8 y/ ?        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).3 }5 x# D* v* v& N0 ^! Z- i3 e4 R
    % b7 Z+ r+ W: u0 B0 S
    Example: Factorial Calculation8 n( H# |# y$ g# n0 r

    & y' E' o- o8 J5 N. R! M% [; A+ SThe 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:
    # e4 _! x' q5 m7 t& m4 m/ U/ c4 {+ a* K" T/ {' A6 l
        Base case: 0! = 1: h, H- K1 \7 g0 S' g' Z
    # z( l' g5 C1 B' @* }
        Recursive case: n! = n * (n-1)!4 }' A, `# h" P. s' O0 s
      [/ w9 T) Y; X2 K6 ^% j- L
    Here’s how it looks in code (Python):
    ! ?- ~7 k  @8 X$ h( a9 [python: R  @, U8 ^* E+ w  v' e
    6 C2 K4 n9 w+ i7 F

    ) U. ?1 [$ ^2 z, N" N, d. \1 Rdef factorial(n):
    5 z* M  M# J4 l" w$ t    # Base case" }, E( h+ l5 v
        if n == 0:* `0 @6 \- |2 Q6 y- Z! v
            return 10 |6 R( x+ M0 m! _; ]& N. r
        # Recursive case0 Q9 S" d# d- P0 w  H/ {4 x
        else:7 j' P& r6 [: Z4 A! }. ~
            return n * factorial(n - 1)
    ! i! M0 b! a3 S, ?' p- g; D$ ?  T' K4 Q! ?* ?) J
    # Example usage0 E% R+ ]9 b/ u- M# Q1 S+ p9 V2 u% W) _
    print(factorial(5))  # Output: 120
    + _% k: I) B4 N9 I/ l/ q' A
    $ e4 i" W) J- h1 A4 d- ]* m8 nHow Recursion Works
    + a( E+ X4 k! C$ K( a2 T0 [3 s3 J/ s9 B1 \  N4 m, W2 ^
        The function keeps calling itself with smaller inputs until it reaches the base case.3 a% e6 D+ E7 [
    ' k2 L4 X) G! f  A, o
        Once the base case is reached, the function starts returning values back up the call stack.
    ' r; q  M4 T  h/ C1 E8 G/ E/ P9 E
        These returned values are combined to produce the final result.
    1 O( f. u! i- Y" W3 ?$ Q6 `
      {2 Q2 v; r) Z4 i5 v6 j. OFor factorial(5):& L7 Z4 ], ^; s% h0 X, F

    ) c% N; c. e+ i9 Q) o% w  ^; c8 G5 I( ~9 D+ f- |' L
    factorial(5) = 5 * factorial(4)
    8 {3 H/ p7 N+ a: A' t" R5 O& @- Hfactorial(4) = 4 * factorial(3)
    1 B' G& {  h. u/ p& D* b$ T5 p* Efactorial(3) = 3 * factorial(2)
    8 u# B) ~) ^/ ], M4 y" sfactorial(2) = 2 * factorial(1)
    9 N$ D3 c6 W& m  D2 j4 tfactorial(1) = 1 * factorial(0)9 ]9 v- t. s% y, l3 p! g1 P& h
    factorial(0) = 1  # Base case
    2 e, w  R, u' D* A; K* p8 Z# F2 h) @8 q2 i
    Then, the results are combined:! q7 Z+ n* _& U+ T: ^/ q2 E

    - H+ G$ n' B. b' J
    4 m- @$ v9 ?4 K( r5 C# J: Pfactorial(1) = 1 * 1 = 1
    - `: n7 ^1 v. C) ifactorial(2) = 2 * 1 = 2! Y5 g- c, k% G) k  y" z
    factorial(3) = 3 * 2 = 6) l. R, X. E0 A3 i
    factorial(4) = 4 * 6 = 24
    6 b. z6 ?, U6 ^0 K. _factorial(5) = 5 * 24 = 120
    % c: W+ i1 R8 o# x$ U
    . N. I; L+ y4 tAdvantages of Recursion2 a/ T: ~& K3 u( l0 x
    ! A" A: [* H! f+ ]6 g& B
        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).0 {) I# B( d$ a( d0 e" X
    . t8 X; K. A% @  m8 ]) B- U+ M
        Readability: Recursive code can be more readable and concise compared to iterative solutions.
    * m5 u, r& n5 a! N" S5 `, e7 |' [- F* z
    Disadvantages of Recursion/ A% p7 [% b, T0 H5 K9 m) j

    6 {7 y$ L0 x# e  r+ P" \' i    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.
    * R% Z( K, }7 c2 x' ^) L# t7 M: C! N6 R+ x9 r) p/ L
        Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
    0 t2 t1 M( V+ A
    6 V6 r5 v1 N6 v' D* HWhen to Use Recursion5 U. [# z8 h3 U* B, S% e
    * w- q2 `% x2 K" t3 w0 e9 P# R0 D
        Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
    ' f0 i- [9 R- A# C) c8 B
    " R1 t1 X  |7 I8 X; H    Problems with a clear base case and recursive case.) t1 z' `: A0 S) k
    3 x. `+ ]& H+ g
    Example: Fibonacci Sequence" @) N( ^+ S1 o# k1 f
    ( R) i% j' L: R
    The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:- V' [, Q  f& X$ G: A$ q8 x4 S
    4 E4 b0 d* }( L+ O! ]2 D( G
        Base case: fib(0) = 0, fib(1) = 1: a8 D8 |' |  M; v0 k$ T
    . c$ }5 K- ]( w/ p' u- s
        Recursive case: fib(n) = fib(n-1) + fib(n-2)
      B+ W( K5 O  D, v2 }* }: S: N) ~  g! {: @% {
    python
    4 G2 K& l; @0 ]) w. m" [9 J# A' H4 ~
      }  m9 d1 m2 K3 }( E9 G/ e2 K$ C: v3 n9 N$ Q6 m% h. ~
    def fibonacci(n):$ y, p( O$ _0 L
        # Base cases5 a' {1 ~" G' e4 ^7 i3 c2 `
        if n == 0:+ i4 S) c8 D5 x' _: E
            return 0
    2 ?5 O5 m4 C/ u( Q- G! I  ^    elif n == 1:8 k& u8 |' U% ~" l5 A8 K( \
            return 1
    8 ?6 l. z, O, ~2 r9 ^4 O- b    # Recursive case
    & J+ z" {9 t+ i- N) \    else:! L1 i+ d1 [" O5 q
            return fibonacci(n - 1) + fibonacci(n - 2)
    " G( ^  a! ?$ i6 W7 W! G6 L( p, D6 l8 B2 W, r3 _+ O
    # Example usage1 A6 c+ j+ |7 S
    print(fibonacci(6))  # Output: 89 a0 J0 w# R2 N  Z( w

    & A* I, w* v% X0 D' q5 dTail Recursion
    # d3 ~/ b( x, X: c! l7 v+ \; N, _+ s
    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).
    ) X, L/ L( \% e+ t
    ' b2 @* o1 ~$ n( S7 W) ?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.
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    地板
    发表于 2025-2-2 00:47:27 | 只看该作者
    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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