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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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    解释的不错
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    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。* _) N4 B+ q5 }

    * V3 G2 Z) ^# Y' c# X% M5 g( | 关键要素, \2 m- ?! S' v. H1 U
    1. **基线条件(Base Case)**$ o1 W4 K, I1 ~" M: d4 ~9 V
       - 递归终止的条件,防止无限循环- D) M1 O) ^. i5 Y, C- c
       - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1
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    9 \' q1 `, r1 e) t2 V* d2. **递归条件(Recursive Case)**4 A: S, m8 q/ Z4 I( j7 x
       - 将原问题分解为更小的子问题" B1 L9 @. q1 W* ?7 h# t6 t
       - 例如:n! = n × (n-1)!
    4 }1 \4 y! @8 y
    4 h, A) b0 q, [. \# G  [+ |; l 经典示例:计算阶乘
    0 ^  s6 [# {$ @( [: M  I$ Qpython1 W, ^! J* y& q
    def factorial(n):
    + \# k+ J. d1 B1 H( L    if n == 0:        # 基线条件6 {7 u( G- l* ?
            return 1
    9 J$ N8 u$ o- D9 Q# }6 S+ l! z    else:             # 递归条件
      j2 q( X$ _. ?# N        return n * factorial(n-1)1 Y- e4 v0 ~" B7 V. S8 q3 y
    执行过程(以计算 3! 为例):
    4 a/ R" J% c  rfactorial(3)
    ( L6 p- L; R6 H9 `. n7 y  a7 c3 * factorial(2)
    6 X$ N7 U: j& |" V  ]& I3 * (2 * factorial(1))
    / d5 M3 j$ k6 Y2 Z) X1 R3 * (2 * (1 * factorial(0))): [; B4 M+ ]' m
    3 * (2 * (1 * 1)) = 6
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    递归思维要点+ K" @1 @) K, F& k" T
    1. **信任递归**:假设子问题已经解决,专注当前层逻辑! E6 j! t, e6 s0 Q& u1 M
    2. **栈结构**:每次调用都会创建新的栈帧(内存空间)) v% h' K8 P7 d5 B. `2 Y! z1 ^
    3. **递推过程**:不断向下分解问题(递)' l; ~! q4 Y& W7 [
    4. **回溯过程**:组合子问题结果返回(归)- m9 s( s9 E# R6 D; c& |0 R
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    注意事项$ M- t+ c; V& t  Y
    必须要有终止条件
    $ q; c! b( ?" h递归深度过大可能导致栈溢出(Python默认递归深度约1000层)6 U0 l( a" c0 G' _( R# K3 ^7 v% f
    某些问题用递归更直观(如树遍历),但效率可能不如迭代- v9 t  a4 Y! l! f) s) r& K
    尾递归优化可以提升效率(但Python不支持)
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    递归 vs 迭代0 l7 [$ v. V+ }) Y4 b$ f0 R8 h
    |          | 递归                          | 迭代               |
    / N9 R+ p4 D* U8 V- S|----------|-----------------------------|------------------|* h, g1 X# H% S2 [: z+ R; Q
    | 实现方式    | 函数自调用                        | 循环结构            |
    + ~9 ]) Q7 k8 ]- u1 c+ W( A; ]) t| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
    * B; r8 U8 F/ u: e" u# s* H| 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         |* l0 n4 ]) i* ~, k& D5 [# M2 o
    | 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |
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    9 h) S3 V, P  F 经典递归应用场景. t0 ~! G8 y9 @/ D
    1. 文件系统遍历(目录树结构)
    2 ~/ E7 O% x1 L( g2. 快速排序/归并排序算法/ e' u7 [6 K6 [& n) _0 A9 V. A& C7 v
    3. 汉诺塔问题* Y8 `. X+ \5 `: a$ ~
    4. 二叉树遍历(前序/中序/后序): _) h/ O+ N6 y4 W3 y" Z
    5. 生成所有可能的组合(回溯算法)( u" R$ c8 E6 A; z
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    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,
    - F0 L% V7 P( q. N我推理机的核心算法应该是二叉树遍历的变种。
    ' U7 k) j$ u- G' a& i8 ^) Q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:/ D: Q5 C& [7 t( d' T7 o- M2 C: Y8 Q
    Key Idea of Recursion0 q3 @& E* k, |1 v7 D. a8 w
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    A recursive function solves a problem by:! r8 o/ e0 m9 W0 H

    ' V1 v8 Z, r  w- I  ]$ V! V    Breaking the problem into smaller instances of the same problem.# x% H2 Z" b5 U& n% @/ c

    / N, i+ U7 f' ^* C8 d3 q3 Z    Solving the smallest instance directly (base case).
    $ c! r% a1 f( p. G0 _' ^4 k8 b# s" _2 M
        Combining the results of smaller instances to solve the larger problem./ L: d1 V/ ^3 o! `  R" G# c0 H$ G
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    Components of a Recursive Function: `, m/ w% b8 W! x

    0 W- @0 K/ r) s- t# h% o1 q; x/ q  a$ h$ n% z    Base Case:& p6 C/ a% n& ]' a7 O5 E$ ]5 {* X
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            This is the simplest, smallest instance of the problem that can be solved directly without further recursion.; B* i5 I/ U8 |# G7 o* X/ {
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            It acts as the stopping condition to prevent infinite recursion.
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            Example: In calculating the factorial of a number, the base case is factorial(0) = 1.
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    8 m& E0 |& v7 G    Recursive Case:: ^5 V3 H' ~5 U  u" _( P
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            This is where the function calls itself with a smaller or simpler version of the problem.; A8 A5 f. E" c. X+ R+ b/ f

    6 t9 O5 l, Z/ g: e" ?; O        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).7 t, i' N5 Z% _8 [

    5 `. I3 {8 S! N: l9 h. B7 DExample: Factorial Calculation
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    ; _2 t8 }  B- y( eThe 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:) ^5 d% b0 ]3 y8 }/ {! r

    : x; a' z3 B0 w9 @# F/ I% _# g    Base case: 0! = 1
    ; d- l& P7 ?, d! m
    $ }" E$ C: O# }( p) s0 F    Recursive case: n! = n * (n-1)!& f6 q: J4 `4 X- y2 U+ _. x* T

    1 i" q. a' {  ^% l; h  tHere’s how it looks in code (Python):
    * i: ?5 x3 L8 P& c4 v# T) B# ipython. x: q. N) a' M; u; a& t1 F0 v

    ) w3 ]0 g5 n% o! w# Q9 r
    8 s; ^$ z& i2 \+ Xdef factorial(n):  J. i4 U+ r2 H5 P) C8 Y4 K
        # Base case# J3 ~8 A$ z5 C# I; @  [
        if n == 0:! k# M9 w* S9 {# w
            return 1
    ( [9 O# F+ P  F( F- I$ |2 u    # Recursive case
      A9 E5 ]2 |: k    else:
    5 \& E+ `0 G& R. a' K1 i        return n * factorial(n - 1)
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    9 S! Y! L8 @5 r1 [0 `# Example usage. P1 Z/ v0 n) p6 U! v. i
    print(factorial(5))  # Output: 120
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    How Recursion Works( l/ O- _: D# \  e" |* l

    0 T' \1 t6 ?: d8 z    The function keeps calling itself with smaller inputs until it reaches the base case.
      W7 H. A" f. S7 j9 B5 w3 ]' L8 ]: n+ i- D+ o+ A
        Once the base case is reached, the function starts returning values back up the call stack.0 X( @, c- F# ?% _9 j1 w6 ]

    9 }- U' y; \9 M! T0 |/ [; p0 Z    These returned values are combined to produce the final result.4 V# [! [' K5 h; m5 ~6 }( @% s

    , P+ h! n# R& L# J- kFor factorial(5):
    . x5 c+ m$ ?- \+ P1 z* h$ D4 d. U  f0 U* v: N) H8 l: ?% T" n/ t
    5 V4 ?- j5 S+ W1 d8 w
    factorial(5) = 5 * factorial(4); {5 ~, x% S8 o: b4 Y2 e3 L8 \
    factorial(4) = 4 * factorial(3)
    0 \1 D  S- @$ B" Kfactorial(3) = 3 * factorial(2)
    0 d7 l3 }, F6 S. M0 E! i  Wfactorial(2) = 2 * factorial(1)% |# q; y% `( G0 I; t
    factorial(1) = 1 * factorial(0)& `% \$ ]; `+ ^- @: ^. x$ `
    factorial(0) = 1  # Base case
    , a* I/ H0 f$ [1 B" s6 Q
    ! o9 s6 I1 M% @4 hThen, the results are combined:) `3 Y4 d( b+ M
    " |! ]" Y# A% b8 \+ h
    - }+ h; z  J. H" @' U6 Q, p
    factorial(1) = 1 * 1 = 1( p3 O# G6 T, `1 g! z* O- w
    factorial(2) = 2 * 1 = 2
    + f- y% I) z6 h" N/ \  Hfactorial(3) = 3 * 2 = 6$ |' d0 x, ^3 x2 l, j& }) {: G5 v2 X8 P
    factorial(4) = 4 * 6 = 24
    6 _  `2 j3 W2 z& G+ A1 r$ @+ k0 afactorial(5) = 5 * 24 = 120
    # e4 C2 H$ @! Y, e5 D6 y- V+ D& l  O! a8 b' q; T4 u
    Advantages of Recursion' g/ D, b3 P3 X' R% s; n

    ! i- D- L- s4 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).8 F: T3 Z6 T# t3 i: n0 z
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        Readability: Recursive code can be more readable and concise compared to iterative solutions.
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    Disadvantages of Recursion
    : u- X! }% V! C  n, d0 ^/ ?/ h* r5 a2 k  G/ e5 r* [! \! C# [
        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.
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    7 m+ `+ Z8 C+ u$ n7 f& V( ^& I    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).% Y7 _- g* ~  G5 `0 u3 u, G0 I
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    When to Use Recursion
    9 Q0 a* P" G8 `' Z% Z/ w3 F- n0 P& @: O# y/ F. S8 W* i
        Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
    1 E8 x: q, [/ w$ ]3 `$ M/ ^0 y! ]
    2 h' A! j2 V- V    Problems with a clear base case and recursive case.% b4 W. H0 `, ]

    4 ~) X( O' d7 l- r- i, JExample: Fibonacci Sequence; W, w4 [3 I5 {* l* Z1 i: n. U: i

    ! \7 d# g8 }: X3 S9 i" LThe Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
    ) o6 T5 a6 B, j/ g" L
    % K5 ^8 N; M: x$ G* e0 V, @; t    Base case: fib(0) = 0, fib(1) = 1
    7 V( A* d9 C# M) L. Z8 U* q
    . L9 `1 D6 |# l8 K5 P- E. z- ?) x* p    Recursive case: fib(n) = fib(n-1) + fib(n-2)4 v0 `) e! s6 B3 C* F( J: i

    ( w/ k; }" r5 i! kpython% X+ s7 o* N7 ^+ p$ s- d
    , n3 h- V4 q7 W5 K' U. X

    2 u9 G# @% @8 |1 \, s# Odef fibonacci(n):# J- o3 a* }% I7 i6 C' w9 a0 \
        # Base cases
    $ S# E! h+ J  B    if n == 0:4 x) [6 f* k0 }# L
            return 0) |% ]: B% x% r. ^
        elif n == 1:
    . h* {$ _" T9 Q* p4 ~        return 1
    ; y0 C$ e% V1 ]- `6 F# n* W    # Recursive case! c  v7 [' T0 L  }! `
        else:
    7 I3 b  y& F' u. |) R        return fibonacci(n - 1) + fibonacci(n - 2)
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    3 c$ \8 n9 B( C+ c+ F5 }4 n) |' \# Example usage" g9 ^2 }" [) A' q% _3 x. M: m: ]
    print(fibonacci(6))  # Output: 8
    5 l, r  L8 {0 l, W: K
    4 x& M! Z: R& UTail Recursion
    % j) k  Y, {/ h( S
    8 V6 m  ~; v2 @) nTail 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).
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    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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