爱吱声

标题: 突然想到让deepseek来解释一下递归 [打印本页]

作者: 密银    时间: 2025-1-29 14:16
标题: 突然想到让deepseek来解释一下递归
本帖最后由 密银 于 2025-1-29 14:19 编辑
) m3 Z' D- `6 Y
5 D0 r7 T6 G) Q: o& A" c解释的不错
/ J1 ?8 E4 t0 D+ P: v" T
  P  n% O5 q4 G  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。
! Q- N$ D$ G: f* x
2 ~& y8 |) A2 l5 j. Z$ X4 h 关键要素
0 }; \" z- x; E7 }" D, D7 o# R: D1. **基线条件(Base Case)**+ S2 q, n& m3 [
   - 递归终止的条件,防止无限循环
# D$ T2 G* {* T, P0 {7 m/ p   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1
1 r7 l# P  J7 K2 ?" s  V' q
9 a. S# i: v+ i. R2. **递归条件(Recursive Case)**
3 e) Y. g- I7 ~. Y4 K   - 将原问题分解为更小的子问题- {& T1 ~- K9 _
   - 例如:n! = n × (n-1)!6 D; Q( D$ a/ |- C
. M5 ~5 X; L- O+ e2 g3 k, Y0 C
经典示例:计算阶乘& x4 l6 i$ C1 z6 c( I- q
python5 ?/ X, s% `0 u) M
def factorial(n):
6 c; B1 o* a) ~6 q7 N7 {' ^    if n == 0:        # 基线条件2 z' _0 R- a1 W) m8 h
        return 1) Q% X& D0 M* V  W$ y/ v$ _
    else:             # 递归条件: r/ z. x: F5 B" Z
        return n * factorial(n-1)% X4 t/ g: |, }5 B
执行过程(以计算 3! 为例):
, O% w) [* {/ d# W: tfactorial(3)
# M& M; c  t! Y4 W0 `3 * factorial(2)
6 L' e, v6 a; ^) M, [$ j. t3 * (2 * factorial(1))7 k/ F/ |" B9 P+ j. |5 p% k% r
3 * (2 * (1 * factorial(0)))) X" _% q" o' A
3 * (2 * (1 * 1)) = 6
' x4 `3 r( c" u# Z' D) j) M8 Q
  p9 H/ G  t) T/ x" S 递归思维要点
' q& B6 D1 M, x1. **信任递归**:假设子问题已经解决,专注当前层逻辑2 Q6 R1 p* }4 k2 d% Y
2. **栈结构**:每次调用都会创建新的栈帧(内存空间), r# n. J: b4 _
3. **递推过程**:不断向下分解问题(递)
% W% I/ J7 j: \. A7 w! y4. **回溯过程**:组合子问题结果返回(归)
3 g$ V( V; v" b( D- S2 K" w; P9 Z
注意事项
* U7 ~- B- a& ]( u  a- x% Y1 H必须要有终止条件7 l* G2 o: R7 S/ p, x2 b( k
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
. W; u) C0 Q4 g+ n8 V某些问题用递归更直观(如树遍历),但效率可能不如迭代
( e7 H8 t7 E/ z尾递归优化可以提升效率(但Python不支持)
7 m, n2 ]4 H% R2 ~* Q6 k  }+ h( M( {$ U
递归 vs 迭代
- W* l% u5 [6 Q( d0 j) q+ [6 ~: Y: `|          | 递归                          | 迭代               |( }4 @* k4 T( }1 u  G
|----------|-----------------------------|------------------|. o- _9 B# t- {5 M( F3 a: A0 Q( n
| 实现方式    | 函数自调用                        | 循环结构            |2 c& l& z3 X! `9 L# C
| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
7 e3 _( P2 N0 f1 B| 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         |( z% W3 I. [) x, c2 x$ t
| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |
7 E. B, d0 d: B1 e+ A5 |& X6 ?3 H9 @, _+ M0 e1 H" o/ y, N
经典递归应用场景7 l; _6 Q" K$ q
1. 文件系统遍历(目录树结构)
2 W. e' _- w+ G2. 快速排序/归并排序算法  K  [8 d5 a3 C, p7 e/ c5 B
3. 汉诺塔问题
3 S" L5 x" u1 Q, v7 @4. 二叉树遍历(前序/中序/后序)8 }, [, e2 Y; V) y, R- J+ S
5. 生成所有可能的组合(回溯算法)
6 I- ?0 v1 H- Z0 y; d. Y  ^" |; W' L& ]4 f+ ^) @' U4 h
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。
作者: testjhy    时间: 2025-1-30 00:07
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,
* P. }7 [+ D6 A) ^6 o6 K我推理机的核心算法应该是二叉树遍历的变种。+ }& o/ ^% [, ]% P1 C) ~
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。
作者: nanimarcus    时间: 2025-2-2 00:45
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:
( {2 c* E$ Z  S4 {, ^Key Idea of Recursion" C/ r. O( d% C8 ], o! [! k" G

3 q. `  l7 h3 Z/ hA recursive function solves a problem by:" v9 _- D0 k; E* h! S

0 z+ _# p5 C& B    Breaking the problem into smaller instances of the same problem.
$ a3 L; g/ {, F6 e
4 j# o0 @) v1 L; a    Solving the smallest instance directly (base case).
5 E2 X% I1 S. d+ \9 a1 K4 O& }3 C9 v3 g- }
    Combining the results of smaller instances to solve the larger problem.+ w  ]+ l5 Q9 d. W

. G- ?7 z& c* eComponents of a Recursive Function, x  X, ^7 C+ X+ R
* Q$ W5 O. J: K( {! J
    Base Case:: K+ I  w# O) N2 [6 Z2 J
" k5 Y7 ]; J+ Z- J; r, T( e
        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.4 v0 U5 y! v, k

& N7 X& G' m6 Z        It acts as the stopping condition to prevent infinite recursion.
' t: l& A  D3 Q
1 W7 O/ y; w! X8 r: F5 }        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.3 A  \3 d# K$ e8 Q

4 q4 Y0 h3 f7 r9 O    Recursive Case:. j- X& d8 k. @& K9 l( [" z6 s
! q  Z9 `2 n! R( b6 F
        This is where the function calls itself with a smaller or simpler version of the problem.
2 }" J" V( _; D  e0 i7 @
# o/ j0 G, u% x. ]" ]        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).
/ w  g: ^' y6 [( E* Y: d) z9 c( S2 L# ^
Example: Factorial Calculation
0 ~4 i# A, k8 _" ]( N. B
2 X$ y* k8 V4 dThe 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:9 E1 s4 r/ \4 m% ?

, H% o! t' m( V. X    Base case: 0! = 1
' E+ j' \' n( c' m
0 a9 z6 [& x7 C+ ]! S% L8 f    Recursive case: n! = n * (n-1)!1 I1 N' M0 E! S8 E4 |

) s4 B" W# s2 y6 u# `Here’s how it looks in code (Python):
* B' r- K$ V% C4 \python- Q# I1 w9 F% ^/ j/ Z: M2 w9 ]8 W
# ^( u8 m3 z1 K$ b2 B, X' ^: g

# v5 m# P2 ~' U" w; v3 ddef factorial(n):1 B3 e8 J6 W" X: `) k: v
    # Base case
0 N! u  T% g, K3 d2 C4 `! M7 _    if n == 0:
7 g; ?. H' g$ v! u        return 1
. {& s/ ~, Z2 a. x    # Recursive case& T, S  c! Q% Q0 c8 R7 S
    else:& m3 Q6 `. Q, g8 t+ a0 j
        return n * factorial(n - 1)/ f4 w( P' \; ^: V0 E9 t- ?
. f+ r5 b& V  Y" q
# Example usage' d6 L8 _/ f6 @4 y/ [$ T
print(factorial(5))  # Output: 120
1 Q$ W3 r  O7 n% o
3 {( C1 r; T5 }. Y" {How Recursion Works
2 @. M) ~5 ]* i) ?' Y3 O) }, [) k. m2 C. K( N5 t$ i
    The function keeps calling itself with smaller inputs until it reaches the base case.
0 y' O9 Q6 c  V" c, y% K5 ?$ w5 j! l8 V  }( l
    Once the base case is reached, the function starts returning values back up the call stack.
& z+ J  ?6 F1 ?1 y& W4 s0 }% }1 M, b9 l" x: |' j
    These returned values are combined to produce the final result.3 @% i% m$ P& V- r2 }
! `, c5 }) w& Y! o$ `# v7 |
For factorial(5):
1 Q* f8 X2 P8 ^: n7 a+ o0 y4 q
( I3 o5 I, J1 s+ r, e
' {9 Q8 @! X% v0 k( h; y, Dfactorial(5) = 5 * factorial(4)  j, i( B& a$ R. i2 ^. {9 Q1 ~
factorial(4) = 4 * factorial(3)8 W/ c/ Z8 G# H; o8 _7 ?9 H
factorial(3) = 3 * factorial(2)
) ]( v" E  \3 Sfactorial(2) = 2 * factorial(1)! E' n* v( B( N$ L  u
factorial(1) = 1 * factorial(0)
* n, _1 Q" Y5 @: v# @factorial(0) = 1  # Base case
1 e8 C% R' r* u( |8 y0 U  {1 Q8 H, C, w! P
Then, the results are combined:
. L' [+ C9 P. o* I$ A1 d0 D* r' F! K" ^7 I

2 F0 a( a- r; E0 Cfactorial(1) = 1 * 1 = 1
. Q7 C( V- \  u% g! sfactorial(2) = 2 * 1 = 2( {0 q4 e7 k) l3 E( T; k
factorial(3) = 3 * 2 = 6
% w0 G8 X4 P; I( y" }3 s! ufactorial(4) = 4 * 6 = 24' f" {4 j8 R; X1 P3 q+ m
factorial(5) = 5 * 24 = 120/ I1 o0 P3 E6 j, W
) `9 x+ l! l& ]
Advantages of Recursion
$ d7 F: Y5 b1 N. F7 l4 y2 u
. T/ G) q. b6 h$ K( f    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).
7 X, {- p' J6 r  K- Q: @0 o
8 A  r. F1 Q3 r7 z/ H+ |    Readability: Recursive code can be more readable and concise compared to iterative solutions.
0 z$ o$ w3 L& d$ h+ G0 G
' [8 A! T' g% d4 c9 ?" a( lDisadvantages of Recursion
+ h# b5 B+ T2 J) k/ E2 L0 i  K
, @3 j, m. H& J0 n$ _9 A3 V  O! m    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! [/ `3 f; Y; q: |$ g, D- T0 U: W  Q/ X' ^  b$ F' S7 E
    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).9 ]1 Q5 L. Z+ {+ }, m

) X$ W3 ^; _8 O$ D6 n% L* zWhen to Use Recursion
1 p2 T* h9 Q8 q+ Z7 _8 d5 j4 q$ k8 V% G# p- R3 k
    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
) ?5 p  X. p5 s; E8 q9 O9 X, n- |* X1 ~
    Problems with a clear base case and recursive case.. q/ L6 X' O+ c7 a) g* p' E
+ [1 {3 ?$ l3 M7 i
Example: Fibonacci Sequence
( L- I/ J" d1 Q* x& m, |+ _
0 f1 P6 Q* r. L: a3 d6 r6 {( EThe Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
" }# i6 N$ [5 `+ n8 K6 h
0 V" Z! [! [8 }0 I9 }4 ?    Base case: fib(0) = 0, fib(1) = 1
5 h! P& q! U* d- M" U3 t$ u- }- R
8 O; R7 ~! k. u- A    Recursive case: fib(n) = fib(n-1) + fib(n-2)
+ f" T/ k4 l/ k& C; z, x. A" s
% b1 a1 f" u4 Ppython
+ \* x. t( c: f( G+ h* w0 [
$ N1 j  D6 }5 R, h; Q, |
2 I2 V6 M6 F& |( d7 d9 f" J2 L1 y) m9 sdef fibonacci(n):" F+ ]8 x( I; }
    # Base cases
* `1 }4 V- m# D    if n == 0:
) G8 `; K3 @6 ^+ J( e' e        return 0
( ?. i8 Y. X3 \4 H* [7 X9 {    elif n == 1:
% U! W0 g! I# v! U: g0 L        return 1
4 e+ F* i2 i, e% P    # Recursive case# t9 {$ p- ^3 {3 `* x. c
    else:
1 R% p$ \& S5 S- [& @, C        return fibonacci(n - 1) + fibonacci(n - 2)
9 a5 L2 k6 G3 ~" ~- `8 w) Y8 p$ u! J: q( Y
# Example usage" d- ^  a. k9 A* B$ O" j/ F* w
print(fibonacci(6))  # Output: 8
, C4 G; b( B( _. i( k3 l
& _9 z/ V% Q0 N6 PTail Recursion
5 R' k7 S$ l) v6 l. A( u$ A, _6 i. D! d, 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).. g! s2 `/ @* N! \, W& n

5 V* [; w% e8 h. N; B5 B+ J0 tIn 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.
作者: nanimarcus    时间: 2025-2-2 00:47
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。




欢迎光临 爱吱声 (http://129.226.69.186/bbs/) Powered by Discuz! X3.2