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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
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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 2 L  W, v2 M- c6 K) m# p
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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。
    8 ^0 {! N. s1 Z$ O( a) e
    & s# g4 X( t) Z3 } 关键要素& l1 U- }7 ?; ?
    1. **基线条件(Base Case)**7 _* B, b4 C2 m/ i
       - 递归终止的条件,防止无限循环
    ) o+ U1 I3 r! S/ b   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1
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    2. **递归条件(Recursive Case)**$ q: X# y3 |# ^& V& ~2 F$ N$ N
       - 将原问题分解为更小的子问题
    " N% o  D2 K# T; d- R& Q   - 例如:n! = n × (n-1)!4 H/ _+ G( q6 r2 h
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    经典示例:计算阶乘$ x5 ^. e( I. F: R& p% _7 u
    python, o6 ?& H& ~: s  b4 o% R% b% L
    def factorial(n):- r- `7 y" d4 c
        if n == 0:        # 基线条件; m8 p  n1 E* u( k* X
            return 1
      u% @# R0 P* b. O    else:             # 递归条件
    5 ^2 C9 D. p5 N        return n * factorial(n-1)! ?* z4 p9 m5 p' t: Y" z
    执行过程(以计算 3! 为例):
    & r" X" w" m+ v; c9 N. f9 f4 ffactorial(3)( M$ C2 N' h% o1 \* K( W# C/ a! u
    3 * factorial(2)* }9 b2 Z5 D( ]; t5 _- T; T0 _' e
    3 * (2 * factorial(1))
    4 L- A! D$ u9 J7 `3 * (2 * (1 * factorial(0)))" B6 s+ M$ d' \3 B; F/ i3 g+ h# X
    3 * (2 * (1 * 1)) = 6  o" b: Q0 _1 ]3 H
    2 D- O0 K" n) [6 I/ F4 g
    递归思维要点5 v  T* D, L$ g- p! s
    1. **信任递归**:假设子问题已经解决,专注当前层逻辑
    9 l. H' O: z; Q9 k% s2 x. W2. **栈结构**:每次调用都会创建新的栈帧(内存空间)8 J/ p: i1 C4 z  l
    3. **递推过程**:不断向下分解问题(递)& j: W7 Z3 o! H5 x
    4. **回溯过程**:组合子问题结果返回(归)
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    注意事项4 u/ f6 O+ R+ }. u
    必须要有终止条件
    , s% }, B. o, N1 f递归深度过大可能导致栈溢出(Python默认递归深度约1000层)0 U$ E" Q2 U& Y; O
    某些问题用递归更直观(如树遍历),但效率可能不如迭代" ^7 u) m# g: {  k. Q0 ^
    尾递归优化可以提升效率(但Python不支持), H2 E: y6 A% |$ z
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    递归 vs 迭代
    & p' e- b* q" l- G# `: h|          | 递归                          | 迭代               |. |! f! |; `  }% I
    |----------|-----------------------------|------------------|& c3 g# h# C. x$ I: p# W; I0 G* y, n
    | 实现方式    | 函数自调用                        | 循环结构            |
    6 a9 b! Z4 i% s( V" v1 @4 P| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
    ; Z0 d3 }# J( ?. L+ k| 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         |  w5 j; m& Y  T- y! L& b
    | 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |
    , c, F% s) {3 {4 p+ L7 C+ U1 O
    / e: U- w" c/ J, V" y7 y9 W4 F7 y 经典递归应用场景
    ' ^: Z8 P# r( L7 W% \8 R) m" H& Q0 g4 Y1. 文件系统遍历(目录树结构)
    : P7 M% J. G7 g% t2. 快速排序/归并排序算法
      f8 Z: @( d* p8 T( }3. 汉诺塔问题5 k& V5 z1 ]0 c$ K
    4. 二叉树遍历(前序/中序/后序)  z8 J) e% T- ^
    5. 生成所有可能的组合(回溯算法)
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    # ?9 j% ?0 ?2 ]% H# 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。

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    1 小时前
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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,1 }# b+ q0 u4 X/ B1 p$ I1 G+ Q# x
    我推理机的核心算法应该是二叉树遍历的变种。/ @: _4 _$ l4 L; Z
    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:
    - H6 n4 m- _. ]: k- Y' SKey Idea of Recursion, {, ~( o9 m) O! U& g: T$ ^" ~
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    A recursive function solves a problem by:8 H. f; Q* C6 @* c& A! _# x8 J" R

    ; f. ]/ @3 [% d) b    Breaking the problem into smaller instances of the same problem.$ _8 i; \6 N: v

    3 E' Z# L/ w) e. _( Q; H    Solving the smallest instance directly (base case).& o# Z) @9 p4 y7 s& M6 g  t

    & Y( p4 }# d. K% [4 x  K    Combining the results of smaller instances to solve the larger problem.( x1 s" }8 l. I' b: c6 v
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    Components of a Recursive Function
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    / U. U) s) [. H, B! m1 m    Base Case:
    8 u7 I) N8 w% z
    3 E8 v( ^$ @5 ?8 I        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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    ' r( Q* X* @7 r5 I7 k: g        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.( {( `8 z1 @( s% W8 J9 g) \

    2 M5 m0 r6 T; x2 A    Recursive Case:
    % |  Q& a5 \/ G4 f  y* F; v/ }/ f6 V5 d! _
            This is where the function calls itself with a smaller or simpler version of the problem.
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    9 @+ y+ m" t" t% q, ]        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).
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    * {2 h* M+ s7 J9 VExample: Factorial Calculation5 o: ~8 B  J. ^

    " Q1 S5 p' L5 h4 P8 m2 CThe 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:
    3 A  }, y# ~$ \7 c/ r
    " U! {3 M% P  {    Base case: 0! = 1
    " |) g0 x: A1 e3 O9 C3 a  ]% Q9 U* j# i
        Recursive case: n! = n * (n-1)!8 p' K$ o4 s+ |9 K+ }4 z
    / u' z) v1 S0 T- e9 B( d- K3 n
    Here’s how it looks in code (Python):
    0 A; i6 A1 ^/ G9 Npython
    ! `" E8 }* x' j4 _
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    & J2 R: _0 x/ i. udef factorial(n):+ |/ |% R9 u4 I& E+ l' U8 d' o- N6 E
        # Base case
    - I1 |/ O5 d+ m# _4 t    if n == 0:6 C) a+ z$ }% i: N3 R) P
            return 1
    6 b9 z( \) K' t    # Recursive case% X" h) t9 E1 L3 S+ e5 p$ k
        else:
    8 @5 U/ K, o8 P        return n * factorial(n - 1)
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    0 e: L5 d+ l" B2 R5 w! ~' S; f" F# Example usage/ x7 Q! E0 O0 p, ]- e1 n$ ?
    print(factorial(5))  # Output: 120* w. K4 {1 i) j
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    How Recursion Works: @# r: |8 d8 Y: b3 w, g
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        The function keeps calling itself with smaller inputs until it reaches the base case.) ?, c" e# q% p& K' O
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        Once the base case is reached, the function starts returning values back up the call stack.1 G) D  ]: F! D/ ]' c# t0 J) z/ Y
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        These returned values are combined to produce the final result." o: q. k/ S$ C' I" ]  t6 F- ^
    ) _% n; _6 T# z8 }' X- u
    For factorial(5):
    8 W) s- K3 k+ b) i, j9 q! N
    " E# ?0 k: c7 z7 k; l7 E8 w; J! R5 }4 p7 f1 s4 F9 h; K& E% K  z# y
    factorial(5) = 5 * factorial(4)
    . U( O; K" {! Bfactorial(4) = 4 * factorial(3)+ E0 f. V. u( M, f8 K. _3 J
    factorial(3) = 3 * factorial(2)
    5 P5 ?$ p/ y0 M: \factorial(2) = 2 * factorial(1)+ |" u1 @0 ^" R  ^
    factorial(1) = 1 * factorial(0)
    % T/ w( d6 L; Y. h- Q7 ^; p; |factorial(0) = 1  # Base case, A; P- R# U) w' W6 R7 C' q+ q; p

    . ?% b/ o$ C( U4 d" E% tThen, the results are combined:( [1 p! n" {( ~! t- u% q
    $ y( g- B9 E- H& d9 w; }' j& x

    3 w1 t0 U6 b$ E' ofactorial(1) = 1 * 1 = 1/ L! m& i! p4 z' ]  c" c& }
    factorial(2) = 2 * 1 = 2
    & w: K7 G$ Y8 E: p8 U5 v  U: \factorial(3) = 3 * 2 = 6) A& e0 B9 T/ E
    factorial(4) = 4 * 6 = 24$ \+ Z2 X2 S7 i
    factorial(5) = 5 * 24 = 120* S3 J) z6 b1 @- r

    0 _7 ]& V! `2 T( }' p# F0 eAdvantages of Recursion  H# Y! s, J) r1 o( T
    ; a3 {6 R7 i5 M8 s4 |8 y
        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).
    $ S9 P8 C# ]; X) S1 Y8 \
    $ T) R. y( {) f% Y4 r    Readability: Recursive code can be more readable and concise compared to iterative solutions.9 q9 K, e$ j% k0 F" g& R8 A2 m

    # E6 x5 h% O0 F4 R# ?" }, vDisadvantages of Recursion* E  N. d5 h$ I
    & Z+ D7 a5 z8 I. B3 D3 K" Y, 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.* D! Z4 e8 _2 F, h4 L; }
    ! n  I: `$ r: C
        Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).+ Y. f/ B* M: R! h2 `

    7 s3 e* f) h' d$ K" _/ sWhen to Use Recursion' A6 S3 }2 N0 q' a/ ~% Y( }2 B

    2 s* s5 t- q# V2 ]1 ~    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
    ! s2 w0 Q8 p: \% W! A
    ! u) `) I& c( K& w+ k% S3 e    Problems with a clear base case and recursive case.( ?# Z7 x- f* }( L& v/ ~& g% @9 H

    ( J% a$ C( x1 V! jExample: Fibonacci Sequence  i2 O4 n7 u* X9 o6 `
    / }) [% Z# {" B" h! F4 i4 k
    The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:: T$ A: c3 \/ f
    ( [4 E' N% Y/ W& M) e$ a6 P; Q- o. Z
        Base case: fib(0) = 0, fib(1) = 1
    * B) Q; a: `/ W
    1 w4 s. y) A: O) G) `    Recursive case: fib(n) = fib(n-1) + fib(n-2)
    " s2 U! q7 ^9 G# Q. N! R* n0 h6 `( M/ s7 E$ q0 x0 ?
    python
    2 [" \5 |* f& X, ~. _; G. r  j/ a" B2 S! F) r# O* s' f

    ! d; w/ l- N' z1 F% q9 K9 K5 ^def fibonacci(n):
    & W8 z" K/ P# \: G7 c, u- k    # Base cases
    * k3 z9 |& k  m. r    if n == 0:
    / e: m" Q: E9 r9 l  V1 B        return 0
    9 J5 j# d. O# `    elif n == 1:# d& v! n6 ~% J0 H* w  w
            return 1( ]; V' ^5 N$ O, S
        # Recursive case/ n7 X1 o! u' b* k
        else:- V6 i: ?5 S$ s! Z. P  L7 b
            return fibonacci(n - 1) + fibonacci(n - 2)4 a# f* V+ q8 v- w3 B
    : [  [  |5 D  I
    # Example usage( ?3 u4 ?; b+ {3 i
    print(fibonacci(6))  # Output: 8
    9 E9 D$ b) I! f  J6 |0 y4 J. c. J- p& r' E/ M) ~% S
    Tail Recursion2 N" H7 Z% O- j4 F! W& e+ I

    . Y# h: E3 m# Y# cTail 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).* g& [' e4 B- x# i; f
      x; V  r, b, n0 c
    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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