设为首页收藏本站

爱吱声

 找回密码
 注册
搜索
查看: 2755|回复: 3
打印 上一主题 下一主题

[科技前沿] 突然想到让deepseek来解释一下递归

[复制链接]
  • TA的每日心情
    开心
    2025-9-8 05:08
  • 签到天数: 3 天

    [LV.2]筑基

    跳转到指定楼层
    楼主
     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 8 Y3 C; c/ c6 l5 F, \1 c: ^

    % e% N3 M3 E- Y4 P解释的不错
    4 F8 _4 U0 Z5 |5 A/ N- X" o0 T# h7 [# m5 Y# Z% R
    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。
    $ x2 Q5 S+ R* T+ c4 h& I
    & B+ F  c) g. } 关键要素
    + R; ^  D* H/ D1. **基线条件(Base Case)**
    ; c7 a( @' L9 j/ o( i1 T1 J   - 递归终止的条件,防止无限循环
    ) V. }  ?* R$ V   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1
    ( Q  q% \2 a  T1 L
    . U7 s% h# z) o5 h* x0 d2. **递归条件(Recursive Case)**/ r& R: d! w4 @  Q
       - 将原问题分解为更小的子问题
    ; l3 r* H( e' @+ w   - 例如:n! = n × (n-1)!% K; {! k% o4 {; L
    + ?7 i2 y: x4 |0 h/ p9 }
    经典示例:计算阶乘
    0 w% h  ]$ x5 d4 M  l- w$ B  wpython
    * v3 I4 j: v  T2 t% w- Zdef factorial(n):
    + `. ~& V6 k) O0 ?+ y  k    if n == 0:        # 基线条件
    5 k  z  F+ k  ^9 V: |4 h        return 1+ c  N. \) p9 @. e! T$ B; T- z
        else:             # 递归条件
    0 F  P0 B" ^3 j; a7 e/ X        return n * factorial(n-1)4 q5 l  W+ c. I! [: O& I
    执行过程(以计算 3! 为例):
    ; P8 j/ X+ c' S$ a5 C+ s. M# nfactorial(3)
    ; T( Q; l* `8 Y) t3 * factorial(2)
    9 D" `- u% n& [" d# z% |' v3 t2 r3 * (2 * factorial(1)): S5 }/ e, @+ v' n; ~
    3 * (2 * (1 * factorial(0)))
    : ?& n+ M. h1 K6 R9 V. u% H# b; F3 * (2 * (1 * 1)) = 6
    . V& N% W% R2 B) B6 P1 ^
    4 @, {: s: N( S0 G: } 递归思维要点
    ' o" x- b# M$ g4 \* ?2 w1. **信任递归**:假设子问题已经解决,专注当前层逻辑, ^" m. H& v; a1 O
    2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
    3 @; J2 @& V  J1 \6 M  s$ O3. **递推过程**:不断向下分解问题(递)
    ! E& h$ d9 |" c+ O9 _. ]& o' h4. **回溯过程**:组合子问题结果返回(归)+ B+ e3 P- @& H, z  L3 U( K7 [* u

    / u# f5 E' i1 C5 E7 M 注意事项
    9 G9 j5 F: q# {/ K; ]% N5 O必须要有终止条件
    4 x4 ]- b+ y0 U8 S7 M2 V+ u递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
    * P2 [  d  S1 u0 o某些问题用递归更直观(如树遍历),但效率可能不如迭代( o9 l8 h7 {3 U- M8 {- J9 I. U# x
    尾递归优化可以提升效率(但Python不支持)  d9 C  p1 w8 ^7 o# f
    0 F8 ^+ L! B% N* ?) c. x
    递归 vs 迭代$ g, t# m' N. n9 C- O5 m
    |          | 递归                          | 迭代               |' N8 J6 {( m: S# ]
    |----------|-----------------------------|------------------|- z( ~( z: r* S6 [1 T% F- o
    | 实现方式    | 函数自调用                        | 循环结构            |# V! z3 L4 t6 S
    | 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
    ! J0 _) S& R# k3 F/ q4 j0 y( y5 `| 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         |
    * D0 ^2 \2 U' b  H# X$ x| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |
    ) m0 j7 w+ n4 E4 o2 j" G! z/ A8 C
    7 c3 D4 n; @7 r$ c* {* K 经典递归应用场景' v; A" X2 }6 [' R8 Q
    1. 文件系统遍历(目录树结构)9 i. J0 I9 \. \6 j+ S
    2. 快速排序/归并排序算法0 f! B( @5 A8 c
    3. 汉诺塔问题
    ( X" `& _* z7 g2 W" m& x3 v! Y4. 二叉树遍历(前序/中序/后序)
    9 E8 L/ D$ e& Y4 Y' x5. 生成所有可能的组合(回溯算法)* M0 m% J" |% [- G6 @# G* L

    8 V5 j9 p6 ^! \+ j2 J5 ?6 N. r& d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。

    评分

    参与人数 3爱元 +26 收起 理由
    pcb + 4
    老票 + 16 涨姿势
    住在乡下 + 6 给力

    查看全部评分

  • TA的每日心情
    难过
    14 小时前
  • 签到天数: 3246 天

    [LV.Master]无

    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,
    2 q  t% Y9 I, K1 n" t' _) Q) T我推理机的核心算法应该是二叉树遍历的变种。  Z( n& A$ Y7 a
    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。
    回复 支持 反对

    使用道具 举报

    该用户从未签到

    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:) @7 {2 z% T  C2 r7 F
    Key Idea of Recursion+ y* S$ c$ U$ ^' G4 R- r1 u4 Z$ p7 U  a" e

    9 @5 f; i- j! I8 G. f+ }- hA recursive function solves a problem by:
    & t: S' o) V: d7 l/ n7 r6 M' Z( ?: v! w1 s  E& y& p5 ~& \
        Breaking the problem into smaller instances of the same problem.
    4 |* y4 w3 `' [6 B) |4 n! V. K8 J
    " f8 g, b. I5 _    Solving the smallest instance directly (base case).. T1 A$ q* V; e# Q  g
    0 l, h! {2 {) i4 [
        Combining the results of smaller instances to solve the larger problem.! v8 h6 z7 g+ g5 T9 l" r: [

    $ [' }5 F% _, bComponents of a Recursive Function3 ^2 e( _; \- y. s
    0 W( e) Z$ ~! _+ B* x
        Base Case:
    & ?6 d" T: n1 \) Y1 N1 ]% B" @
    : z$ N) ]. @9 N  a8 Z7 D        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.8 g+ a9 B! `- A: _8 X/ x$ f

    , d- e1 N- `6 y; i0 q        It acts as the stopping condition to prevent infinite recursion.& F2 z0 k0 Y% `( u7 x

    - x2 q5 c, N' W  e2 c        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.
    ) z* D+ O, B2 c+ K% b. f: r4 C, _( y6 a
        Recursive Case:
    6 A" R" e: V+ I( V
    + f  K) G- }! J" N% Z8 S$ ?        This is where the function calls itself with a smaller or simpler version of the problem.4 c( P8 E  w, p- R% l  f9 @

    9 R  R" _0 C" k& J        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).% x# S1 }6 I; G
    & r4 X* {' @# F0 V( Y
    Example: Factorial Calculation
    6 F$ @& j: t4 y5 ~7 R$ [+ b' P4 F" V1 j% h, a
    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:
    ! i$ |4 P; J; f! l" Y* e4 I3 W9 t
    + O5 b$ C$ P8 ?  j* I8 H% o    Base case: 0! = 1/ |. G$ a9 I+ R, D' f( |" ~2 t$ M
    ! c1 M. T& w3 m8 v
        Recursive case: n! = n * (n-1)!
    : o" N8 y" N0 C& N8 O# p! o# j, u0 C" Y
    Here’s how it looks in code (Python):
    4 P+ O. y. H4 g9 @python# y9 f/ x, J% G( K$ @) \, n6 U) S

      z/ o( P* d7 n8 V6 M7 k/ a, A* S. B
    def factorial(n):9 l  y: m4 t0 K- z
        # Base case, i/ n" r6 F9 ]9 r
        if n == 0:
    5 Q- f- ]: v/ a/ u1 Q% u8 O5 [, y) |, h        return 1
    0 [" N1 y: {7 T5 B* \    # Recursive case- W; ^0 V- E5 B1 H0 d! b, l
        else:+ E* z0 x3 q  U3 a: T' T- u; D& j
            return n * factorial(n - 1)
    / c3 h- c# e& C7 \
    # @2 m3 L0 k! }! [8 S# Example usage
    8 J* q, W+ r$ K( V  e+ uprint(factorial(5))  # Output: 1202 D* X' d3 `/ w( G, P
    / o- G* D- ]1 Q& X$ [0 `
    How Recursion Works$ A' _0 \0 A6 e+ {# X

    7 G: G' Z6 \) i8 B6 D* n    The function keeps calling itself with smaller inputs until it reaches the base case.5 _! z- b8 C; U& F/ r; ?4 ^4 {3 l

    0 i2 W0 Q) v5 R1 g5 }  h    Once the base case is reached, the function starts returning values back up the call stack.
    % U! u: W6 {7 ?, R3 i3 H6 d! I/ e+ s: R& m4 I; s
        These returned values are combined to produce the final result.( k0 x$ z* g( B% G8 l

    2 _3 }5 a; L) m9 \: a* ^0 @For factorial(5):. a1 d' ^5 v7 r6 ~# t0 G0 h! ]# D9 M
    & u1 W, j' Q, q2 S! E1 u; R! f( c

    ' m: G' e0 A3 a- k$ rfactorial(5) = 5 * factorial(4)
    7 O& C/ P+ N( I. N0 `/ n, vfactorial(4) = 4 * factorial(3)
    % u$ T+ C1 p- O) {% lfactorial(3) = 3 * factorial(2)
    + h) {5 b2 X) O9 o4 Z0 D. k0 ~9 ~: vfactorial(2) = 2 * factorial(1)& a, Y6 S5 y. a/ i
    factorial(1) = 1 * factorial(0)
    , T% P' H2 @6 x: z# L4 }4 Pfactorial(0) = 1  # Base case- K# x2 v8 Y4 G9 F
    4 F6 N$ Z5 n1 N/ L1 H! V
    Then, the results are combined:1 w# p$ L; w+ c: a
    3 E" |* Q' ]. \" F/ Q  E" |

    : i8 t9 ]6 \  \- f& mfactorial(1) = 1 * 1 = 1
    8 v4 Z6 ^  V) h9 j% zfactorial(2) = 2 * 1 = 2" V5 o% `4 B& G' N
    factorial(3) = 3 * 2 = 6
    ' I1 c8 U2 q% l1 N" r; X7 s0 ~factorial(4) = 4 * 6 = 24
    . P9 S1 Z0 b0 ?' p: T1 D4 _# a( Nfactorial(5) = 5 * 24 = 120
    : Y+ W0 n% G1 B" i& h  d$ {" f" f$ T0 N6 w' `
    Advantages of Recursion1 z1 {9 a: }8 K) s5 K$ L8 d
    % w4 N; ~4 E2 x# G; Q7 R
        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).
      Z& ^7 v6 B) n& f  _# |
    . T. _8 V- w2 \    Readability: Recursive code can be more readable and concise compared to iterative solutions.
    : y& t. {% `% C1 D0 v5 I* X
    3 I$ X: s: \6 m8 oDisadvantages of Recursion6 q2 Q0 m( i& {4 @- T# P

    ) T. V# j  V4 g; L  t1 i+ ?5 q4 l    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.# O* d; `7 h  |5 ~4 @) ]3 y1 y

    - ?  E& Y5 O* {6 O9 [$ _3 b    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
    ' J  a% Q1 @$ p8 n0 m. p' q; K2 f1 S0 u2 s+ A6 Z. {+ ~
    When to Use Recursion
    ) M. J& F, `# H. `1 h5 a6 j  w8 A" H! {3 i8 T
        Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
    . @) p- H  ?% F, J$ x. h
    $ t3 w) y! d: y, p: d9 g* [    Problems with a clear base case and recursive case.
    6 _" T7 G- T2 q% O  B4 j  G* O
    & q4 s8 k' k: _+ tExample: Fibonacci Sequence
    : `! k% Z4 z$ C, b9 a* m: D2 B
    ( J; x7 h! k3 @) j( Z2 vThe Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
    5 ^: n! x* K( K8 U% m/ E. r* m# c. k3 C& u9 x% J
        Base case: fib(0) = 0, fib(1) = 1; x, N3 N2 `1 ^+ Q* L% a7 ~& ^

    & g. X4 I& F" c  P/ [$ f    Recursive case: fib(n) = fib(n-1) + fib(n-2)
    - J4 F. {# Q( |4 ~2 r, j- x4 ]# O
    : p) C, T4 C: e* `python
    7 u4 p  I# x6 T1 l; }9 S# e% I9 y5 N! h4 k1 h
    5 w! N; g* V/ q0 x/ Y
    def fibonacci(n):5 r: C' U$ V; `' ?9 c
        # Base cases
    " c( k* b: J% U0 g  e    if n == 0:
    0 y. V+ G  p# o% v8 e1 E  ]5 J        return 0
    " U" @$ s. x7 @+ d# Q    elif n == 1:2 T. f7 \! o6 G. U- P& c* u
            return 1
    ! A8 S- ^# z5 N5 t) Q$ L% l    # Recursive case% }' J. l7 S3 N9 D/ C
        else:$ f+ o+ _, p$ v2 H
            return fibonacci(n - 1) + fibonacci(n - 2)8 @- S# a% {# G3 \. P* ^* e# z/ C
    & L# F6 M: ~) K" w5 }3 z
    # Example usage3 ?9 Y: z. k: \1 s
    print(fibonacci(6))  # Output: 8+ m# m. @4 {. i1 T3 o, V
    " q+ q7 N. i* k+ F% q9 O% d
    Tail Recursion
    ! L6 _9 Y! X, u* k3 Z. [
    3 X5 ^3 U4 ?2 f, u7 v4 m% MTail 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).
    ( \4 k1 u+ F5 }
    ; z' U( j4 n- w3 n/ fIn 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.
    回复 支持 反对

    使用道具 举报

    该用户从未签到

    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。
    回复 支持 反对

    使用道具 举报

    手机版|小黑屋|Archiver|网站错误报告|爱吱声   

    GMT+8, 2026-5-22 21:52 , Processed in 0.068917 second(s), 18 queries , Gzip On.

    Powered by Discuz! X3.2

    © 2001-2013 Comsenz Inc.

    快速回复 返回顶部 返回列表