设为首页收藏本站

爱吱声

 找回密码
 注册
搜索
查看: 1322|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 4 }& E  D0 O9 i8 ], d- P& m& a  s
    7 W( X/ \' z2 V8 k" l
    解释的不错: @* w* P/ y0 @' O% ^) o
    0 [8 ?8 X) C4 a# 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。' Q2 G* r, d& b+ c* T1 ?# z

    6 O6 d  y! T' d4 f( @) x6 J5 ` 关键要素
    3 h# d) L7 \" c, C$ l1. **基线条件(Base Case)**! f5 K3 {+ o8 ~9 j! W
       - 递归终止的条件,防止无限循环
    + H+ N6 W/ H. i$ E6 _0 C   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 17 ?: q0 f  U0 h+ R! R7 r6 L

      `: ]0 A' k4 ]0 h5 i2. **递归条件(Recursive Case)**
    - ?" p% O7 n2 R0 z   - 将原问题分解为更小的子问题
    6 s' m- q% m: J9 e' h2 s   - 例如:n! = n × (n-1)!' Y6 o9 U1 u+ Q) c& e9 |' [

    ; \2 A: T: e! f( ~) i1 A 经典示例:计算阶乘8 G2 R3 L+ C6 C
    python
    5 m5 E% x6 `2 I( vdef factorial(n):
    & H7 T6 t5 e6 @5 A    if n == 0:        # 基线条件
    + n9 a" S# [* m' H9 D* ]( `+ H        return 1
    & s. P8 `( H2 l  C8 z4 X    else:             # 递归条件
    1 C6 N" ~# z3 `        return n * factorial(n-1)
    , J0 f, t( Y2 X; @执行过程(以计算 3! 为例):  F" l% t7 N1 |) D3 _% G
    factorial(3)$ Q4 C  [/ _; D9 L! O5 T: ^3 ~' y
    3 * factorial(2)2 B# k' |9 V1 _! S
    3 * (2 * factorial(1))/ w' F0 o) x0 x- m: c/ w
    3 * (2 * (1 * factorial(0)))
    # I* [6 H. ?4 Q5 |0 q( q3 * (2 * (1 * 1)) = 6
    . u+ `) H/ e) ?& w; D) H
    1 P' }0 b# v& Q 递归思维要点
    " q2 o9 A) k: P( n. A, D& k1. **信任递归**:假设子问题已经解决,专注当前层逻辑# s8 K, N$ V% a
    2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
    3 D2 Z0 @; o0 q2 u# h+ s; w3. **递推过程**:不断向下分解问题(递)
    0 q# J/ a% q9 B4. **回溯过程**:组合子问题结果返回(归); P, R8 ^% ~4 i- B1 X2 \& ^" l
    , _0 z6 C2 E$ V* `+ F4 Q* j
    注意事项3 V- O2 N3 c- z0 ~& N/ z; w0 d* t$ O/ |
    必须要有终止条件
    : U+ s2 m1 K/ w递归深度过大可能导致栈溢出(Python默认递归深度约1000层)3 P/ N% P, q* L3 Y0 y4 r0 ]+ u
    某些问题用递归更直观(如树遍历),但效率可能不如迭代
      s& i8 V8 X. _, B; d$ i尾递归优化可以提升效率(但Python不支持)
    9 k  V3 ]$ Q8 g$ H; L; C1 ^- _% z
    # I4 w; f( O- d 递归 vs 迭代
    ) |* q0 M7 J6 T0 V9 X8 i) D|          | 递归                          | 迭代               |" Y% H5 v# W9 s# D- b, M5 F( }: a1 T
    |----------|-----------------------------|------------------|
    # S' I% g+ g: I" Z| 实现方式    | 函数自调用                        | 循环结构            |8 \6 r! H8 H: k( n& k
    | 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |& ]5 b! H( K9 c  N; r
    | 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         |
    - ^; L( v4 H7 i| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |
    7 o0 ~: x. ]% P: G* r7 r' O& ^
    % L' x1 d: J7 S8 a 经典递归应用场景
    0 ^) _1 |$ ?4 j1. 文件系统遍历(目录树结构)/ k1 d9 H7 R. _8 r( A, Z0 n
    2. 快速排序/归并排序算法# f3 u' y$ O* _! m6 l
    3. 汉诺塔问题
    ; ~9 q# B( S: C: k5 J: B5 K4. 二叉树遍历(前序/中序/后序)
    " y0 h+ u# `6 f) _) `5. 生成所有可能的组合(回溯算法)/ a: I; |& `5 d5 W2 u" V) O/ e) Z
    4 f" _6 {: {+ N% c& ^
    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
    昨天 12:19
  • 签到天数: 3107 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,
    # J1 E% h% B- ]我推理机的核心算法应该是二叉树遍历的变种。4 {* p) Z! q; i' k) E$ 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:* @5 \5 {' Y% |: B: N
    Key Idea of Recursion
    . S$ h* m  T! `8 N8 V
    5 e. e' j. ~' o1 z# q: _, sA recursive function solves a problem by:
    ) d5 o, H* L8 g9 j' A" ~% i
    + A# d' @" Z1 E, g; A- O9 R    Breaking the problem into smaller instances of the same problem.5 z! h5 V- k0 ?7 c
    3 d& j0 j; [# E9 {1 k2 V
        Solving the smallest instance directly (base case).9 _2 J8 c) B+ W# \2 s- O% w
    ; B5 R% Q) v( }; T3 F
        Combining the results of smaller instances to solve the larger problem.& Z6 i3 P- r' d0 I
    ) \4 w5 m! e# G% @7 U8 n( ?! U: x
    Components of a Recursive Function* B$ S& K$ r0 I; U+ V5 A

    + ?- U- U, g) r) `3 t$ O! _    Base Case:
    9 [& z  X' G7 C2 L4 t0 h4 j% A: E* _7 D$ X" n' I: t5 W
            This is the simplest, smallest instance of the problem that can be solved directly without further recursion.
    6 W  G6 z4 S1 S
    # @$ p0 h$ x$ ?) }: P, H        It acts as the stopping condition to prevent infinite recursion., X! H! @( p( T# l4 h

    - p; }6 s5 W; t# k        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.1 _1 m4 ], ^# I( t0 J

    ' d5 A2 v. N9 m; k% M    Recursive Case:8 H$ F6 `0 y8 |

    1 E7 ~. z$ ?" k. _        This is where the function calls itself with a smaller or simpler version of the problem.
    7 j# q7 X; w: t- {/ Y1 E+ g+ t
    1 d  l( q' l' p        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).; L) q( c, T- f& F6 f

    7 h5 U3 j3 v3 V7 i7 g+ w; AExample: Factorial Calculation) \: W4 V6 N' Y/ X, f# W) K

    ! c# D; F; b  t/ |* rThe 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:
    6 n- ]+ x% @; h. v; x
    ! K2 C  X$ U" m8 D3 f$ }6 U' w    Base case: 0! = 18 B4 M* j7 `9 i6 y6 T' F

    " d" C6 ]3 ~7 \* D% [% E( w! x( D    Recursive case: n! = n * (n-1)!1 I5 W- b) d6 ^# W' K. _
    + A% E2 f7 I: l9 z9 g
    Here’s how it looks in code (Python):
    9 z/ ^6 D* t8 y7 R% ?- x( Vpython
    & ^% Q# x# v$ X( R9 P4 F1 i
    & J$ j* n* {- h) ?/ I$ w* ~* h9 e8 X6 C) F4 n3 [
    def factorial(n):2 F8 J: H' y$ d0 l6 F
        # Base case! ?: l& X) O" X; p# e# X" p: r
        if n == 0:3 N, W  y  ~% P$ i
            return 1% v2 `8 A" w; w+ z( a7 X
        # Recursive case1 g0 {: [- Q3 C
        else:$ [8 ~3 F' Q0 r* s% `$ A8 A. _' {
            return n * factorial(n - 1)& X3 i& q3 j# X5 T$ _9 _9 F
    : g! i; k- K: M# V7 {
    # Example usage
    ( d0 j& |3 U  S# j& t8 pprint(factorial(5))  # Output: 120
    8 {3 C. p; l! v3 v, f
    4 V! }. K2 D  d. SHow Recursion Works- G4 j1 f* c/ {; u7 X/ G
    5 E2 D, \+ i6 R# R
        The function keeps calling itself with smaller inputs until it reaches the base case.
    * ?: r% I, m+ p* y- S6 g
    # M, N  h- M4 l2 W$ M0 r: i    Once the base case is reached, the function starts returning values back up the call stack.2 F' E# G$ }! }% x

    . {5 O/ G7 T( W# W  A    These returned values are combined to produce the final result.% K( ~8 m/ C) s! e, K, W3 ~  b

    3 l5 X7 V# o$ f" PFor factorial(5):
    - s2 F' \" W* C. {$ v2 s) `- t+ |
    - V9 Y$ z* A) \: V
    factorial(5) = 5 * factorial(4)
    & j; x3 V8 n2 g5 i6 d7 w/ Gfactorial(4) = 4 * factorial(3)- E/ r' F* }9 u% _1 F) H
    factorial(3) = 3 * factorial(2)( b- `3 o' l6 [
    factorial(2) = 2 * factorial(1)
      w4 ]" j& S) H( m5 o: Vfactorial(1) = 1 * factorial(0)
    5 h+ D7 I. c; h) ?% K4 e4 bfactorial(0) = 1  # Base case
    # Q* p$ ^$ @& a1 d( S) h- l6 i+ v: S# ?. z2 P1 {# _! M
    Then, the results are combined:4 ^2 y! @( R) A* f: N  w. M

    4 r' r# j9 b3 l4 }* v. T/ L0 _& M$ B$ P3 N
    factorial(1) = 1 * 1 = 1
    / V- r+ N# q: w- U0 T0 sfactorial(2) = 2 * 1 = 24 I$ c: L& n# S% z4 i1 e1 j
    factorial(3) = 3 * 2 = 6! [; D( }; u) X
    factorial(4) = 4 * 6 = 24
    * [: s  e, ?, C& w8 Q* J3 d% m% S* ^factorial(5) = 5 * 24 = 120
    / q" y! A" T$ F9 B! i" y
    ( t7 r2 X8 t% {! J' \Advantages of Recursion
    - x! |8 a" k/ n& `7 I6 P2 O0 _& F, q1 I. u! I5 M2 X
        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).0 `9 b) _, x" E: V1 k" E5 O  L) G% ^' l
    0 k& F1 |5 h5 {/ u8 \6 e3 P
        Readability: Recursive code can be more readable and concise compared to iterative solutions.) x5 }7 t  `- W7 b8 l
    ) p9 R$ O) k/ X7 I0 S
    Disadvantages of Recursion
    1 B# I1 K2 N6 A( P: P9 K( f3 @& P+ `! u( G1 G! A5 N
        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.
    6 w7 ^# M. |* w4 I5 F
    , L; [) Q4 ]- P( c3 l% N% P    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).) s6 r7 k  t1 n% P
    . s, f" Q) H& A& Q( O
    When to Use Recursion
    0 G' Q3 f2 ^: u5 n+ K! Z( u! g+ v# Q6 H& ~/ x7 w1 }
        Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
      a3 q) y. G4 T: d2 X" h2 F6 w, [
    - n# r5 M" E2 J    Problems with a clear base case and recursive case.
    * A2 T3 N7 Y. a0 t1 _) {& G2 e7 F8 ^! b1 O: V8 P7 i1 \9 }6 ?0 L
    Example: Fibonacci Sequence
    : S  `2 F! e  G/ Z% m4 s: k9 b$ _! K& X+ H, j. P1 ~* ?
    The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
    3 q. ?) u; f: n. C) d% [6 v" S' G& O/ y/ G9 \3 u% G3 P+ ?/ `  f
        Base case: fib(0) = 0, fib(1) = 11 |) E! w* Z( X+ t

    . _6 ~5 g- A/ @0 O) l    Recursive case: fib(n) = fib(n-1) + fib(n-2)
    ' d9 g4 U1 Q, F& {( g9 u- [/ r! ^4 C& m$ s# S9 Y) a
    python
    : d$ m- P# ^) H7 j6 m" |
    " X* F& p' A# ]2 n& b/ e9 }# f- U1 g" _8 `0 K$ B+ s  `
    def fibonacci(n):5 e4 W1 A) l4 }: l+ |
        # Base cases
    ' b( \+ N0 w1 B5 j; i    if n == 0:5 A( E# y3 D7 E) ~; O$ ], e0 V
            return 0( N1 C2 L" {) }. \5 j- B& h8 Q
        elif n == 1:1 x' P9 s4 Q$ ~9 A; g% J+ @3 ?: q# u
            return 1' I5 U, S+ w; J5 V- @9 e8 L
        # Recursive case
    ' _& I8 S5 E4 E4 F- I6 }    else:) v6 `" L' R6 p6 C8 A/ X
            return fibonacci(n - 1) + fibonacci(n - 2)
    ) M: e% d' Z2 a+ u* I3 l
    8 P2 e* y3 h: I3 M3 k* ]9 @# Example usage9 ^) c1 l- H# y! f3 U- x8 d" X
    print(fibonacci(6))  # Output: 8: q; k; a+ ^  U; @2 E* f

      }) B+ F5 t1 Z1 R7 }- KTail Recursion
    " A. y4 Q+ l% p( b' l3 I( Q8 j" {: I& ]0 ]. d1 d# C# k! q& y
    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).
    : @) N. v( |* e8 U6 `( u# B0 T3 m4 l  G( v
    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.
    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, 2025-12-4 10:02 , Processed in 0.037263 second(s), 19 queries , Gzip On.

    Powered by Discuz! X3.2

    © 2001-2013 Comsenz Inc.

    快速回复 返回顶部 返回列表