设为首页收藏本站

爱吱声

 找回密码
 注册
搜索
查看: 2365|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 : P( t/ i& v/ y  W- ~" e

    8 \  r- C+ p% A% P8 Q解释的不错# X1 m5 \- k: J% m
    3 f( a8 C$ z( U0 K3 U
    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。
    + Y' v- [% v% R, r# g
      ~# E& C. L) W7 w 关键要素8 ?  g( m/ s& s: R7 D! h3 o
    1. **基线条件(Base Case)**1 M/ c% t; e6 }5 A
       - 递归终止的条件,防止无限循环
      F; Q* w  m- I1 ?, M   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1
    7 p3 Q$ ]. _; X/ W- Z& l/ O( s" o; M: U0 p% C% F
    2. **递归条件(Recursive Case)**
    3 D( @$ x, R; P0 {! q. ?! ]3 i   - 将原问题分解为更小的子问题
    : r2 P5 L( \+ }0 [   - 例如:n! = n × (n-1)!
    1 _: s/ S8 g% f. d( U6 V4 V- m$ V& ]1 {5 l/ p# _: ]. U
    经典示例:计算阶乘
    ( G3 l0 y5 k! m% T: i% Y+ Hpython& e1 \; N2 F# U) J/ ?
    def factorial(n):* k7 G: g: t& z
        if n == 0:        # 基线条件
    + y7 x  n& v& f        return 1% [1 ]/ z6 j( Z8 p
        else:             # 递归条件
    . }; l- F8 X* d: ~) k* Z& r4 n5 x; D  A        return n * factorial(n-1)
    , I! U- z3 o+ `: [& L% s执行过程(以计算 3! 为例):! u9 p4 h& U' s/ v+ n+ s4 `
    factorial(3)
    3 k: E5 D9 T  K3 * factorial(2)
    " b3 b( }2 h0 R& p3 * (2 * factorial(1))
    / a7 Q  |' k4 R  z- J0 o3 * (2 * (1 * factorial(0)))
    6 W% L  Z3 S4 Y5 Q8 h3 * (2 * (1 * 1)) = 67 [8 d- M3 s2 r, |0 B- J$ W
    ! X/ Y1 b3 B) d2 D4 z$ }
    递归思维要点
    6 q; z3 D; H) [: |0 j  }. g  \1 w0 S  z1. **信任递归**:假设子问题已经解决,专注当前层逻辑4 P# Q6 L7 O7 K
    2. **栈结构**:每次调用都会创建新的栈帧(内存空间)& B" a" P7 @9 L, P
    3. **递推过程**:不断向下分解问题(递)
    # h0 L9 ^' x) C( [/ x! W4. **回溯过程**:组合子问题结果返回(归)
    ( q! \+ e& {% Y8 ~& G5 @: t" u5 E* Y0 Z% s: T( M3 [
    注意事项8 X0 q% M1 d4 ~. J5 ]
    必须要有终止条件
    % u" O6 @4 w/ T) b递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
    & F% X8 z# Q+ |. @# o, ~某些问题用递归更直观(如树遍历),但效率可能不如迭代
    + s4 B" w3 p/ |- E尾递归优化可以提升效率(但Python不支持)
    ! O2 y' \. r  p- ]
    ! d  t5 M& D4 O/ l2 j 递归 vs 迭代: a" J. L: I9 H
    |          | 递归                          | 迭代               |. X, j, r' n3 Q5 i# c" A$ ~
    |----------|-----------------------------|------------------|
      M: w8 Z) Y' I* F$ u| 实现方式    | 函数自调用                        | 循环结构            |% o5 G! ]0 r; M% m3 O& S
    | 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |$ v# }6 V# u) L
    | 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+ @3 G% U; f0 w$ h6 A3 ^+ ^| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |
    1 s$ z9 Q- U2 R4 `2 C, Y# E( z3 J5 m$ I
    ( b  ?# C- k3 [2 Q' V# O7 R 经典递归应用场景
    * W; g4 p6 G/ G" ~! p0 ^1. 文件系统遍历(目录树结构): T7 z* b; p( Q3 z
    2. 快速排序/归并排序算法
    - E, U' p4 V1 f$ I3. 汉诺塔问题
    ; h0 j7 c7 b3 Q9 S& Y6 u- p4. 二叉树遍历(前序/中序/后序)3 `% l/ l7 u. ^/ c" n
    5. 生成所有可能的组合(回溯算法)
    - H% W8 m1 N# i8 [7 |( D
    % r6 l4 v  t8 y/ }0 j3 f4 i$ M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
    郁闷
    5 天前
  • 签到天数: 3208 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,
    * k$ Y- b1 Z0 A8 T8 S/ P我推理机的核心算法应该是二叉树遍历的变种。2 K7 p5 n8 W' e
    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:' k3 l1 d/ z; D9 g8 o1 U8 g3 s
    Key Idea of Recursion0 |) m2 k7 X+ D4 u" x; P

    - g/ \& M6 X# `! T; S8 M# _A recursive function solves a problem by:
    8 L- A9 z" z8 m1 X. |
    % u5 K' l$ {- f! ], {' I    Breaking the problem into smaller instances of the same problem.
    2 R# Y) d3 e# a$ ^
    + d) Q9 \1 Y/ V' |7 R    Solving the smallest instance directly (base case).
    ; X7 p5 i) E) }" Y3 s5 x$ t9 ^3 E6 t& ]/ }5 d3 w5 z
        Combining the results of smaller instances to solve the larger problem.8 M/ r" {  F: Q# r) p( T' M/ k6 k

    % ?, j8 o7 G; M/ m9 VComponents of a Recursive Function
    + P& L) T. J* @# f$ p/ Q, J+ q- d$ C& X  z
        Base Case:  Z& d& l1 \/ [4 }3 l9 z. i3 u- d
    ) E( J$ A4 k: F# I
            This is the simplest, smallest instance of the problem that can be solved directly without further recursion.6 y5 a+ g- F1 t3 E: `/ Q# |
    6 ?) ~  N' v& F/ z" @4 J5 u
            It acts as the stopping condition to prevent infinite recursion.
    * |. g7 n: y+ D* E. N3 E# K; u& O9 \% v4 X
            Example: In calculating the factorial of a number, the base case is factorial(0) = 1.0 \) \7 g( b% S0 O. L- G
    ; _: v& S6 h8 S( J$ `  h
        Recursive Case:
    - P  R) f$ U  k% s# B& O7 `- Q- [- w$ u4 `6 `  @) f
            This is where the function calls itself with a smaller or simpler version of the problem.. \5 J/ N5 g5 K- ~0 ]
    ' Z/ ~1 _/ U5 C. a/ G$ _) n
            Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).& r9 u5 M+ L: \0 b. u8 D  b
    0 F0 E0 O- I) K
    Example: Factorial Calculation
    4 s/ N3 r5 j6 v' U0 m3 }0 ?* f6 w* s3 b+ J( P
    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:
    / E+ R* m3 b) D: g  O
    ( y- M0 k# y& H    Base case: 0! = 1
    ( [/ Q; E1 L. t, ~: f& l: }. c7 D3 s$ g$ J( E
        Recursive case: n! = n * (n-1)!
    - Y' T1 r* w$ Z( M6 h& Q5 I. S, f  b+ M4 B5 x
    Here’s how it looks in code (Python):
    : L( `3 Z" B# u2 G; ?! v& s  ^python: }+ ]5 _2 ^# O
    / P; t! Y9 [6 \

    . C- E; `) G6 o8 V8 Zdef factorial(n):* e& j, i, {; @! Z3 k
        # Base case
    - y! y; [# e" D! l  D    if n == 0:: K% {0 r# l9 W3 |* s' R
            return 12 j* i4 j2 N# L& V' k
        # Recursive case6 a( e8 L+ X2 m: a' O
        else:# G, u% a" U4 Y" \) @
            return n * factorial(n - 1)
    ) }& o1 v9 P6 s/ r  w, M
    2 Y, |$ h2 L/ R! T9 l# Example usage: t: p, k  W$ ?7 U8 X& G2 n
    print(factorial(5))  # Output: 120
      T; N' K& d6 j& |; i. \& S
    2 t  `4 e+ ~( I% }- |$ t3 `How Recursion Works  n* c0 I# u* v7 D8 D4 @

    0 D7 c. r; H* I* y4 f    The function keeps calling itself with smaller inputs until it reaches the base case.
    ) `) {8 e* }4 }# i( s% M3 H2 X+ m/ [0 b
        Once the base case is reached, the function starts returning values back up the call stack.
    7 \/ d6 F( z( n5 i4 x) t1 w  F' l8 a2 A' n6 ?. p
        These returned values are combined to produce the final result.
    - @7 a0 ^" H7 F" z% W! o3 k, v8 }6 L. ~+ A" j
    For factorial(5):
    1 p1 `, j; n# z% }3 f' L$ G$ O+ o$ l' h/ a) E7 Z1 U7 m/ c0 i+ k+ d

    / K& C0 w+ _; I% h$ Yfactorial(5) = 5 * factorial(4)
    ; S: S/ C* C# H1 h* g. _factorial(4) = 4 * factorial(3)7 r* q: V0 x" J
    factorial(3) = 3 * factorial(2)
    7 E/ G( ^3 ^/ {9 o6 o, _4 Lfactorial(2) = 2 * factorial(1)
    ; ]% S+ t/ F; [factorial(1) = 1 * factorial(0)0 h' X' o* Z+ N. o
    factorial(0) = 1  # Base case
    ( h) R- q# f0 Y2 q1 @, j6 J# J# ?1 s: x' n
    Then, the results are combined:
    2 }! v8 F( r% v7 B, D
    " {$ g0 r* U' m& L' T3 G  X7 e3 B
    1 G" a" \$ h( h" lfactorial(1) = 1 * 1 = 1
    ( H7 }8 ~% y* m* f0 _" F# p+ D/ f; pfactorial(2) = 2 * 1 = 29 ~9 G$ }2 n1 [" W% A1 {6 `
    factorial(3) = 3 * 2 = 6
    5 D6 ^8 ?2 h) `7 Ufactorial(4) = 4 * 6 = 24
    ' a+ T' m) p* rfactorial(5) = 5 * 24 = 1208 b! u  [  d/ `9 s0 c, i) R

    " o# H+ B# R: b$ b1 U1 ]4 n/ YAdvantages of Recursion
    " r" d. c& W& i  a4 |. ^# E- J* S2 t# g
        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).8 d$ G: Q: g4 E2 k! q2 B0 S+ B

    ( `$ V# z0 z' O& h' ]! n/ ~9 K$ m( D0 {    Readability: Recursive code can be more readable and concise compared to iterative solutions.  }: ?! A+ x0 U- \

      t  n3 C1 K0 t# L7 SDisadvantages of Recursion  p4 K3 M6 Q0 ^. ?( Y2 Q7 e2 F) _
    - l( u6 A5 e5 E
        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.0 v4 l" q6 y6 n8 n. d* k
    : r5 Y% w5 {9 w4 I# b' e1 g% x
        Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
    2 X9 H& Q+ {" c6 l3 S' ~: @  i0 n/ K; n$ p& {* N
    When to Use Recursion
    + u4 T' M# I; R) y$ g3 ?6 F3 v) b% j9 }0 p
        Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
    # J' q% f3 {* Y: h0 @. Y6 C* |2 p( |2 I/ U0 C" w- [3 Z
        Problems with a clear base case and recursive case.( T3 x: w+ ~6 I" p: C

    / i3 |- h0 ?& q1 s0 }# r, cExample: Fibonacci Sequence
    8 M2 u8 r7 p6 a& d6 y2 r4 x4 n0 g- o. y9 ^) k
    The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:& y- v, @1 G8 T  P
    ( M) X/ N% t% c; v0 X" ~
        Base case: fib(0) = 0, fib(1) = 1
    3 p, T8 u# Y/ j" b3 {$ u8 c! }7 g( j0 f4 X% H6 J2 E+ T! [5 U
        Recursive case: fib(n) = fib(n-1) + fib(n-2)( E8 }( G- k8 H+ a0 S

    & j" T* E3 P, e; i* E0 opython$ l2 L( V4 A& |, f) ]3 S

      |, x. s0 t: G: n& d3 J" R. k# h" C
    3 @' V0 s8 K: y  H+ S! y9 I# e8 gdef fibonacci(n):9 q+ ~8 `* D- ]. k2 |( E
        # Base cases
    6 B! K* R* ]% T# A    if n == 0:6 U  k- H- N. s2 O
            return 03 {9 k5 Y* V9 b# V
        elif n == 1:" I5 H1 K3 W2 W- F
            return 1" R4 y1 O  K! c* N6 H4 Z
        # Recursive case/ S  y% S. ?6 @; n; q2 r$ I$ f, a
        else:0 P0 g# d* K  V1 S- T; D' D% G: ?# y1 r
            return fibonacci(n - 1) + fibonacci(n - 2): _5 @1 |) ?7 y
    $ l+ d2 ]+ a& N+ G5 ?3 P
    # Example usage
    $ {4 y& ~( w7 E; i% N0 v- @print(fibonacci(6))  # Output: 8
    # d% y/ A9 h/ g9 v3 y
    , u9 k/ H" c0 o$ _3 N2 N% OTail Recursion
    + Y+ z+ ^. D7 L) v0 `0 p& j
    : [  k, S; d0 U1 b, k+ |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).
    ) v3 C/ Z9 l" N+ G3 v- O9 H& d
    5 ~5 G( a# `% O0 \' S% B3 Z* VIn 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, 2026-3-28 19:52 , Processed in 0.055059 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