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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
      a3 a! ^/ n0 D8 v# P: x' S5 _# \4 O5 V! P" ^) u. G7 I
    解释的不错0 E! ^& C+ v1 Q3 I/ [+ X5 Y3 K& g
    ! E% t$ {0 X& o0 b8 t
    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。
    / B3 ]+ w7 a" k; ~! A8 f9 j# b3 B  h1 X. j8 R
    关键要素/ e: ^) C5 p' b* U
    1. **基线条件(Base Case)**; c. w$ a' O. w+ t6 E0 U
       - 递归终止的条件,防止无限循环. }. U5 e' v: \8 e( |! B, N
       - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1+ R) E$ r( R# V5 Y6 y, p; R

    * J6 `! D2 z0 f. \! [- Q0 R: W2. **递归条件(Recursive Case)**% Z1 j" V" ?$ o) O1 U
       - 将原问题分解为更小的子问题
    % j1 l) d3 C- S- N% E5 Y2 t   - 例如:n! = n × (n-1)!
    0 l' R/ w% b# ]% w, m- F& ~6 a( O8 R; L( o* {; ]5 L! e
    经典示例:计算阶乘
    . H% a9 c+ h1 ]9 D" Y+ [python
    4 P  j  c7 f% Ndef factorial(n):" x# w& y3 G% T& N9 ~& H! x
        if n == 0:        # 基线条件0 i; A% S9 K+ {! Q( U
            return 1: u( {) ^  i  n5 ]& U& f4 j/ |
        else:             # 递归条件
      G/ X& a6 H9 `8 b( T' `; o  h$ A( D        return n * factorial(n-1)( j# }' D# p$ M! t# I4 W) D
    执行过程(以计算 3! 为例):
    7 l! n, l+ @& m/ a) O) o8 N! r& g! {factorial(3)
    ( u& n) n( V9 o& s! D* o3 * factorial(2)
    3 g1 d3 h  \+ M7 X+ D" X3 * (2 * factorial(1))8 @$ i0 a4 y# ?& Q  y" D# E
    3 * (2 * (1 * factorial(0))), Y" C% B2 x; a3 h, G8 a
    3 * (2 * (1 * 1)) = 61 D& s) Y# @" S8 u

    7 t" k* O& ]0 @( A2 A 递归思维要点$ q9 R3 D" |4 I
    1. **信任递归**:假设子问题已经解决,专注当前层逻辑9 c* e* y. m( z( V# J
    2. **栈结构**:每次调用都会创建新的栈帧(内存空间)9 \) X+ T! U! r- b" p! l" M, u
    3. **递推过程**:不断向下分解问题(递)
    4 c5 F% e# l% [1 y8 V9 w4. **回溯过程**:组合子问题结果返回(归)" Q; Y! |  b* q( o$ I* A, {' E3 v

    9 V9 I) H8 z& x, C: C, K 注意事项+ G8 ]' g( }" e1 ~
    必须要有终止条件# Q! w: n  q0 w- ?# ]
    递归深度过大可能导致栈溢出(Python默认递归深度约1000层)5 t4 ]) h; F$ l- d$ |8 g
    某些问题用递归更直观(如树遍历),但效率可能不如迭代
    9 Z( `( I9 u$ I4 k尾递归优化可以提升效率(但Python不支持), m* |1 z  p( n3 |" v0 [) Q% l

    2 m0 {/ Y' E, h% I/ y 递归 vs 迭代. {0 J- `) Q& N
    |          | 递归                          | 迭代               |7 m$ W7 E+ g0 _0 p
    |----------|-----------------------------|------------------|0 u2 f! Y8 N( y: p( @
    | 实现方式    | 函数自调用                        | 循环结构            |6 f, Z2 S) M$ v  v; T0 E  V
    | 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
    ) d% J* n2 B. n  b: N| 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         |; _3 j" i/ O2 h
    | 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |
    & X  Q$ [; C: f( a
    - o; P, Z9 B# m8 h, O. W8 N1 n+ @4 U 经典递归应用场景
    % w" W2 Y9 j1 |- p- c; c. Q9 r$ E3 L* N1. 文件系统遍历(目录树结构)2 i1 C- b1 q  P; y$ z+ w
    2. 快速排序/归并排序算法
    " l, ~. C2 n) l6 T% F  c  Q3. 汉诺塔问题$ s7 M7 B0 P- R
    4. 二叉树遍历(前序/中序/后序)) J7 A+ m2 x- Q+ K* U/ C( l5 y
    5. 生成所有可能的组合(回溯算法)8 k1 F( f1 k: Y, [+ A. ~% ^

    # i8 p0 C2 w( K/ O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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    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,
    * Q( F5 r4 @  Q  U+ Q1 V3 E+ f5 ^我推理机的核心算法应该是二叉树遍历的变种。
    % P1 g5 y- S) w" u) j# R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:
    ) e3 G9 r, z' RKey Idea of Recursion: ]4 o! p6 a$ i7 i1 a) y8 X$ V3 M

    ; m7 c5 t  D  M# A9 ?! KA recursive function solves a problem by:
    % z7 Q3 |1 L* f  w. h  w( j  o8 D" N1 s) [5 W
        Breaking the problem into smaller instances of the same problem.! h: W! C1 U0 m

    4 W" Q1 l& |& J3 x4 \    Solving the smallest instance directly (base case).
    ! F* W+ M2 j' Q8 O4 M. ]0 Y6 [+ p! E+ y/ K
        Combining the results of smaller instances to solve the larger problem.0 A" }: j3 n2 s* C/ {4 n  h9 N0 S

    5 R3 w0 h+ ~+ ?9 V/ XComponents of a Recursive Function
    5 q1 P$ R. y8 L) a6 A3 w
    , y2 g1 U6 t9 a# k; d% F# i: [/ x) H    Base Case:- ^. U. m/ v1 A+ c* w% j( s
    1 Q" }7 }3 L/ d. l* ?* w4 X
            This is the simplest, smallest instance of the problem that can be solved directly without further recursion.
    ) E7 Y! Q( I( N3 W) G
    ; k6 M6 B/ B8 ^$ H$ ?" Z        It acts as the stopping condition to prevent infinite recursion.
    0 O* y- ?: z: D; }9 V; A1 }. D( P0 y" C- G0 {
            Example: In calculating the factorial of a number, the base case is factorial(0) = 1.2 y* F# b5 |0 u5 B# L! }% J

    - a' ~9 a- Z, {# Y    Recursive Case:
    2 b0 ~3 x# f2 d0 I( i7 R
    : N9 C9 G; d  R' o8 ?; \        This is where the function calls itself with a smaller or simpler version of the problem.
    / i2 \+ {. a2 A" u8 a  C/ F3 F# F: h/ _. G
            Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).9 ?; A7 K7 f5 O0 m; L" t6 {  }! q
    0 b0 v6 W  ~6 {5 ~1 E5 P
    Example: Factorial Calculation
    ; P0 Y0 n) |  @; O2 @3 w  P! m+ I
      Z% L' w* R2 `+ U" F& MThe 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:
    . v- \( {7 w7 c2 {' d$ C. m& G. F) U, U' s& q1 \3 k8 Q
        Base case: 0! = 1( M" ^9 Z, Q1 P3 e5 `
    * v# o( p1 h) p
        Recursive case: n! = n * (n-1)!
    % A% b8 v6 k3 a- @
    1 \0 O& D% O: {0 r  fHere’s how it looks in code (Python):
    + K  n+ S5 Z' ~$ W& H4 |: Upython
    # R  K7 o3 a2 p8 p' O  s" X5 @8 n0 w. [
    ( j7 }3 ~) x1 U6 j9 C& v8 k: o: w3 W" d
    def factorial(n):$ ]- K! o9 e: i0 ^5 h
        # Base case2 A' W3 P, Y2 k8 U. U% F
        if n == 0:  ^3 N8 [6 \( ^6 u1 d
            return 1
    - o7 t% Y$ r4 b4 G* f    # Recursive case. Z/ n$ {9 g& `' @
        else:
    2 s0 ]+ \+ C. J. _        return n * factorial(n - 1)% p- F+ |. B/ [4 P, |+ \5 \

    & ~+ I& F: o6 r: Y! e# Example usage
    6 i. G5 g; c6 c+ s, yprint(factorial(5))  # Output: 120, t, V# s: Z$ W- ]
    0 b" C+ ]9 I9 h: u
    How Recursion Works
    ) K1 H1 ?* Q$ D
    5 J+ Q2 ~# T* V    The function keeps calling itself with smaller inputs until it reaches the base case.
    0 A6 r9 m' Y  M- C+ t
    ' l: p9 n1 ?, L% z+ g    Once the base case is reached, the function starts returning values back up the call stack.
    6 L* C" F3 k$ T; ^/ Q
    - D3 K1 k  y  f1 z    These returned values are combined to produce the final result.
    # e) n" F$ v: K" f  j0 [2 ^+ V4 c2 J" r0 D1 m
    For factorial(5):
    / X: ?( _) W# z$ ~: r
    ! O% }5 k8 B( B0 a3 D% ]* o% c
    / r3 K3 q/ [* G) k. _  p0 Vfactorial(5) = 5 * factorial(4)
    % d- _# q' ?1 ]5 i! n: Sfactorial(4) = 4 * factorial(3)
    ! U% b' d& V$ U6 v. dfactorial(3) = 3 * factorial(2)7 f" C4 U: A9 ]3 `# A
    factorial(2) = 2 * factorial(1): P; r; _8 L8 q* U. Y
    factorial(1) = 1 * factorial(0)
      s1 [! i9 b0 [factorial(0) = 1  # Base case( l: W* J6 `! p* Q
    3 y& v" Q& a  u* C& ~0 O+ k( k% V
    Then, the results are combined:# G8 E; `' x0 \* g
    - D$ N$ ~6 L3 V7 ^3 B) i; {

    + Z' C3 d" q1 [0 }& vfactorial(1) = 1 * 1 = 1
    - Q7 y: b% H2 p6 j9 M' S+ c- ]' @% T& Ofactorial(2) = 2 * 1 = 2% h' y' J2 P' x- n9 F
    factorial(3) = 3 * 2 = 64 _" g+ r. }" C& F
    factorial(4) = 4 * 6 = 24. H6 I" f0 o- o3 @2 s& Q, ~% M8 h4 j& W
    factorial(5) = 5 * 24 = 120) V1 q$ x3 ?7 x" r( s. s

    3 ]  L  l/ S9 s. a, ~8 Y6 R7 h0 UAdvantages of Recursion3 \+ D% N4 u8 H& R. J5 `- G

    1 ]3 p" C; G4 \    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).
    $ `& J; ~+ h. ^0 A
    ) u0 ?/ ?3 f' C! p    Readability: Recursive code can be more readable and concise compared to iterative solutions.
    : i3 b, H& d2 m" G8 ]5 J" [
    & g" F3 _+ o7 cDisadvantages of Recursion
    - H4 _& g5 ]$ _3 `) u* f6 k& ~" U% j
        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.
    3 y# Z6 ?6 m+ C6 Z# w* x" Y1 H/ l3 H5 Y$ l& N% ^. p
        Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
    * A2 u. [; }. A5 Q8 [# e
    & }- m$ u0 S7 D! k# h; F- K6 [' P7 wWhen to Use Recursion' ]' l% K+ P6 i+ n

    ! V: t. b) N: H% Y( c/ w+ ~# v* I    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
    # j, |) z3 h/ r8 U2 j. {: t- ]/ Z# |, j$ [3 U
        Problems with a clear base case and recursive case.
    - n$ f# }- O! q* o/ _( K$ }7 u3 ~. I* ^( Y- L
    Example: Fibonacci Sequence
    1 B$ _! T+ _( d6 l" i. S; Q) j- i1 y+ A" c4 z
    The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
    3 V( h; q2 ^# D4 O. y3 \, W9 }, J; C. s
        Base case: fib(0) = 0, fib(1) = 1
    $ v6 H! ], B* |( t, C$ A: q; }
    ! j' h% I  s9 q$ S' B    Recursive case: fib(n) = fib(n-1) + fib(n-2)/ @' T% Y: F% X2 m. V$ |

    & _) a. B; ~: K0 i  A3 A: S: dpython: @, S8 S8 J6 v% F# `4 C$ i
    " h' d1 ~& _0 L6 W) _4 e
    ! e0 ]/ Z: ~; t6 a
    def fibonacci(n):% Y5 _2 L% K: p1 ~; ^; P* O, L
        # Base cases
    - i5 z; p2 ?: P: F5 D! A    if n == 0:, g9 ^& ^( k  P$ z( N% d' ^+ `
            return 0
    $ o+ K) Y& b3 w5 T    elif n == 1:
    + i& C4 n  {  x) T% i8 S7 i2 u        return 1
    0 t6 |8 X& x4 K0 N; [* P+ n    # Recursive case. n3 s7 P$ U# c, u/ Y
        else:, \$ g) X( x" r: J2 u
            return fibonacci(n - 1) + fibonacci(n - 2)
    7 o" A9 ~" M. \7 V
    ) b: P# y! C0 G8 k9 o  P6 d# Example usage  s: P3 ~  `/ Q2 R
    print(fibonacci(6))  # Output: 8" R: i- A& p- R/ o# g
    ' s9 T: d! c8 m+ s5 z
    Tail Recursion
      O+ y5 L4 ?3 A2 N( N4 P6 ?! E1 Z6 ?0 x; a1 g3 v3 G4 Z0 W' c
    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).9 [: I: L) C; f) O( l& W
    # b( m3 ?# A$ s' W6 B  p4 f
    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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