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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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    2025-9-8 05:08
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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
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    3 S2 N7 n  ]! g, R2 ^, \解释的不错; t9 x# ^7 Y$ ]  E' v

    ) m- K* s* p8 v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。. I7 F( E+ |/ C- R

      n- i6 ]3 s$ C; T: s" ^ 关键要素
    0 P4 M3 I3 B, i1. **基线条件(Base Case)**% ^; @8 r) z/ y7 R
       - 递归终止的条件,防止无限循环. O# J3 z- p% g- \
       - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1% C) P. P, S0 y
    , U! S6 H) r4 ?4 w( g0 k6 p
    2. **递归条件(Recursive Case)**5 Q4 z7 B; l* s5 M  }: P6 p8 @6 c
       - 将原问题分解为更小的子问题" G1 {3 K% H" \7 x( Y
       - 例如:n! = n × (n-1)!8 X, c0 h$ M/ y( [3 I/ Q

    ( J6 k$ F3 {8 R2 \, K% `: x  N 经典示例:计算阶乘
    ) [8 A2 h+ J: Xpython
    . }! O. p- A/ @( J7 U* y6 l+ K; ~def factorial(n):
    / d# ?9 h( p9 W    if n == 0:        # 基线条件
    ; M8 G1 F- G9 ^* O        return 10 f, l; ]- v: e6 O# C- e' R9 q6 t
        else:             # 递归条件  t0 J) E* n/ u% d0 D
            return n * factorial(n-1)! A7 H( P2 I6 f+ g6 S$ Y5 K- i
    执行过程(以计算 3! 为例):
    ( w$ H5 _6 e* e) F" k3 l% u  pfactorial(3)
    + U" `* Y9 l+ S1 F0 {5 a$ B3 * factorial(2)
    ) I+ C5 b. c$ X# q2 V/ f: u: d3 * (2 * factorial(1)); p& A9 @/ ?; g/ m, H
    3 * (2 * (1 * factorial(0)))) o; W: g! J: T% P2 t- b
    3 * (2 * (1 * 1)) = 6% _' S4 e, X- k5 H( V

    ' W: s9 c9 N. O. m. i3 r 递归思维要点
    7 f# ~. ^5 f& k& @1. **信任递归**:假设子问题已经解决,专注当前层逻辑  ?/ q1 X( T/ n7 x3 p
    2. **栈结构**:每次调用都会创建新的栈帧(内存空间), M% g$ r$ Z# y& b7 `
    3. **递推过程**:不断向下分解问题(递)
    % ^) w3 e# v' X' d' \$ a7 [' Y1 o4. **回溯过程**:组合子问题结果返回(归)3 P6 A' Z* T6 X/ C! U

      k1 K: b1 q* ]* [ 注意事项
    1 k+ `5 r/ x- W9 ~0 A& E必须要有终止条件! u2 B2 v. K6 C. ^* n3 }+ O7 J
    递归深度过大可能导致栈溢出(Python默认递归深度约1000层)$ \# t0 _3 ~5 j" `- t/ a' w( W
    某些问题用递归更直观(如树遍历),但效率可能不如迭代
    / a- l. r! `+ H尾递归优化可以提升效率(但Python不支持)7 _6 O& D1 E' t' K: n
    ! E7 w( {+ d3 l$ z
    递归 vs 迭代
    ; ^" S$ p$ @$ g3 p8 O( K, F: z5 X|          | 递归                          | 迭代               |
    , X8 P$ R# m, C/ J$ C& g: Z|----------|-----------------------------|------------------|
    ; r( b+ S" |( k| 实现方式    | 函数自调用                        | 循环结构            |. e, ]# O5 M. b- I# d7 t0 \0 C# N: e9 \
    | 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |( C( L& }; H" V9 N7 t( ~5 \
    | 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         |
    ( ~( }/ g% h4 b1 M* F1 C" v| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |( D; R# d+ m+ R5 f* p1 q
    ' s- ^2 E  V3 d# R/ C
    经典递归应用场景) m% u. \+ m4 U7 |0 Q
    1. 文件系统遍历(目录树结构)
    ! M: l+ b9 d& r! N2. 快速排序/归并排序算法
    . n  ]# y& E0 h5 W. Y* y$ A/ E  [. a* L3. 汉诺塔问题
    4 ~5 ~0 w7 }% }3 e4. 二叉树遍历(前序/中序/后序)
    5 z$ ^4 O) Y9 N( p. \/ C$ m5. 生成所有可能的组合(回溯算法)% f; M; \) e* |$ f% M
    ( v" W: L" B" q0 ]. Z7 U
    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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    昨天 08:45
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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,
    % S9 C. P- N  A2 K  z  @$ c  Z我推理机的核心算法应该是二叉树遍历的变种。  v; M$ r/ x0 f, Q& ~- `+ ?. v
    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:
    7 ?$ f8 x+ X( R4 M% {4 R( yKey Idea of Recursion8 x0 l9 F+ z% z' k+ k' m% Y

    % V3 U( o3 T- [A recursive function solves a problem by:7 Z; X- |, j) m: L. e$ o
    - n2 p5 X7 D4 C
        Breaking the problem into smaller instances of the same problem.
    9 y2 h' |5 u  q( r" n& B# _% |" V5 _) l( c$ x
        Solving the smallest instance directly (base case).
    % w# Y% m- G. Y
    , u$ d( e/ a2 B) W/ E0 D) `8 {    Combining the results of smaller instances to solve the larger problem.
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    & b% T! Y) c4 v* XComponents of a Recursive Function
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        Base Case:
    5 s9 q& J! J/ \; `6 J% s9 y1 s; q, e5 B5 C9 H9 O3 e
            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.
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    / q* u- D: R" ^9 _) [3 ~8 M# Z        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.
    + T& P  a  |% v# X; x5 P0 }, P4 Z' ~+ T# V
        Recursive Case:
    2 h; k' ]! b' c3 S+ B. D0 E" d0 L5 W2 s/ l' a& V1 J; n, V! a: X
            This is where the function calls itself with a smaller or simpler version of the problem.# i, f, s+ G0 C/ Z
    " E+ r+ `, i+ R9 M/ t6 O
            Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).
    4 k; Q4 }2 P( X: K' W3 ?9 V2 k" X& o7 ?0 @& [* n. |4 T6 u
    Example: Factorial Calculation
    : q! L9 ]% @7 f" w' b4 b2 r- {2 K1 s  k8 G/ I6 W( ]" J" Z, _/ j
    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:  B- Z# l& q* \9 G
    1 L; I2 Q' U1 C
        Base case: 0! = 10 Y* P, i+ \" d/ B

    ; s/ V" Q( D' x1 ?+ {    Recursive case: n! = n * (n-1)!7 g6 }& O) `! S7 f

    5 W" u. h% z- ]: @) c, SHere’s how it looks in code (Python):
      D& z) x$ e2 G' d1 opython
    8 f* \5 D/ D( G- P& N! w/ E" G: {) _5 S  _7 [
    7 r" q3 B7 g# J
    def factorial(n):8 K& O* }4 u. Y8 N3 p7 G0 O; \
        # Base case2 B$ e+ M) V5 K8 Y* ]/ u7 f9 J
        if n == 0:
    ( F' a2 s) j- T3 Z! W        return 1! n5 ]2 a2 c( V0 k! w! F0 g
        # Recursive case: C. f& u2 F* p+ m/ ^% M
        else:
    : ]1 t2 _+ X5 B3 v/ ?        return n * factorial(n - 1)
    ! i( ], [0 a, @6 a$ V" Q9 _' O0 j% b+ I  W
    # Example usage
    ; K9 C7 Y$ j! eprint(factorial(5))  # Output: 120
    9 k! i0 T" ]4 z) x" E8 T4 ?2 z0 M% c; G+ q3 }( R  Z! v
    How Recursion Works
    # O$ b8 i" ?; w& S: w3 X9 e9 }- G* Q5 v
        The function keeps calling itself with smaller inputs until it reaches the base case.
    9 `( A" A$ q. ]% J
    3 M" _1 p, z4 W/ E3 r    Once the base case is reached, the function starts returning values back up the call stack.
    4 |. V$ e5 k" o% K
    ( a( b& ]: o0 Y+ i+ a; C    These returned values are combined to produce the final result.* k; t; g# T+ }! `9 m4 H; K

    : w$ q4 Y8 b4 t! P0 g9 p! R8 o% z+ RFor factorial(5):4 p: c1 }# V/ x) g1 @+ e% J. V, \
      i- g& r3 A2 a. T
    # @9 C$ c3 h( D  O. V. u" p
    factorial(5) = 5 * factorial(4)6 _9 h1 b$ b% o& i
    factorial(4) = 4 * factorial(3)) P% a' X) B$ b: Z( n2 X6 t
    factorial(3) = 3 * factorial(2)7 B0 k  D( M7 A6 I
    factorial(2) = 2 * factorial(1)
    6 V3 {+ d0 D( dfactorial(1) = 1 * factorial(0), |# r# S% H& h
    factorial(0) = 1  # Base case# R4 F/ j' r9 z

    4 c4 j' {4 C6 J7 T3 s: iThen, the results are combined:
    & n  N' F- V1 u& L; e
    . T$ A! Z3 G. @) ]
    * i% _3 Z4 b: Q8 I9 Z# V. l1 z8 Cfactorial(1) = 1 * 1 = 1
    # o1 \5 `1 ~/ k! G. Hfactorial(2) = 2 * 1 = 2% Z$ {; e4 Z6 u' W$ H8 E
    factorial(3) = 3 * 2 = 6* d9 j3 X5 N4 p4 y. U) c8 P
    factorial(4) = 4 * 6 = 24
    ; H; [6 H5 i* }) J: A3 Ofactorial(5) = 5 * 24 = 1208 j7 H. Q" h5 d2 G% r

    5 a; D0 }+ r- a( `: ?6 N% C) ~Advantages of Recursion! F: ]+ N, R- K
    ) Y" ~" X5 U! y" h$ Y9 H
        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).
    % y( j  o: z+ z( G
    ( o' P, R; N) c, _& [7 E    Readability: Recursive code can be more readable and concise compared to iterative solutions.
    6 z# W; l' V# x" D+ s! F: X+ ]& d
    + P3 f3 U# B4 f# p& j+ {4 _" _3 X' RDisadvantages of Recursion
    $ I% I" _  O: x2 d; q! Y9 K3 Z' M  H9 X: g0 C) y
        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.. }+ k$ a& u+ N. L; ?" W: `
    9 w9 _" d6 s( \' i
        Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
    : Z  z0 O/ [) H/ B3 G0 j2 w% W9 S5 _' [
    When to Use Recursion9 {' I6 \4 ?2 X3 `  Y, A
    ; k" [2 K8 h9 y& L3 A0 ?
        Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).! J! M0 z  P) o7 h: S, X; E/ l

    , r# T+ J+ p/ L; }3 P( o# B    Problems with a clear base case and recursive case.
    : a8 T7 ?+ c* \
    $ n, K* p. P/ A6 v' @+ UExample: Fibonacci Sequence8 }1 w' v4 J, s

    : ]$ ]& @$ r+ H2 w7 M5 L( v# ?The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:* x( h% D: u  T, w

    5 J4 ?2 D' E, V. M4 i" {    Base case: fib(0) = 0, fib(1) = 1
    $ J0 H# K. a  Z, N5 }! S! V( W- t2 H" `8 i. w# d1 O9 Z
        Recursive case: fib(n) = fib(n-1) + fib(n-2)5 n# K" \% O8 T1 ^' n) D4 Q2 o

    1 Z" v/ A6 i, a9 r$ Rpython
    . D0 |! q' B' \) m" K2 ^8 w: y" c3 Z6 S1 }
    " q! k, p' O, A% ~! X
    def fibonacci(n):
    $ J6 T2 H% R- u5 ?0 O6 o' D) ~( ]    # Base cases0 |6 _9 O0 z6 k) _- ?
        if n == 0:
    8 W/ q# l# @, r% U% [1 W        return 0
    : Z* D3 T' Y7 @& [( ^2 o9 X, N! d    elif n == 1:
    ( I9 U7 V& P  S/ ~6 i( \) A' s: t        return 12 X; j+ k, o9 J2 T! o' G
        # Recursive case6 m5 Q8 Y: G% J6 j- v
        else:0 ^$ d9 S) D6 q  g
            return fibonacci(n - 1) + fibonacci(n - 2)- f+ A# t$ H( k+ ]
    % J0 ^# U' t( D4 p7 P  C  m6 O, A4 d* W
    # Example usage, u" C6 B  E( v& ?2 k
    print(fibonacci(6))  # Output: 8
    / b( g# h6 T& R% ^. B% z
    4 ?# ?% A( _) ~Tail Recursion
    7 ?6 u9 F8 S% L/ j  v5 o! F- E* T. M. ^. i5 e0 ^6 n: _) }
    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).; F- A1 U. W; v
    2 l3 l4 ?# m4 w$ d/ u
    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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