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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
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解释的不错1 f& u. Z$ L  `3 r8 W- J% F
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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。0 ]; j/ M# ?' Q; ^* d  d" q
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关键要素
- S. f1 u, e6 ~' ?' S! w9 s2 y( s1. **基线条件(Base Case)**  G% W: [+ z9 W# n: ]; |
   - 递归终止的条件,防止无限循环: V& q5 d6 q: W! }# s  @
   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1
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( ]+ |2 w, \, l+ B2 p8 p6 d2. **递归条件(Recursive Case)**2 {, u- x  s2 _0 ~7 y3 k+ n
   - 将原问题分解为更小的子问题: U: r5 n" g( N8 @  x
   - 例如:n! = n × (n-1)!
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# d4 m. o0 k' i7 z 经典示例:计算阶乘; [; B# t0 b& ^5 `* s3 j
python8 H: q! A, f& U9 t# h& h! z
def factorial(n):8 w; B6 z% M$ b3 A: ]
    if n == 0:        # 基线条件
3 x* W- M: @) a/ y        return 1
1 x: k8 p! n( [& P( h    else:             # 递归条件
. K* R+ y5 A# p/ Y1 w        return n * factorial(n-1)& z1 v! a' B# J% ]. C: \3 @/ N- }
执行过程(以计算 3! 为例):
* b( D9 w7 o6 Yfactorial(3)
* ?5 }7 c1 M8 @7 c! _/ B3 * factorial(2)
4 K3 y) C7 I2 Q) z3 * (2 * factorial(1))
  s1 I8 ^7 R9 I+ ^# g3 * (2 * (1 * factorial(0)))
; x" _7 p* b0 G8 _3 W3 * (2 * (1 * 1)) = 6, g5 `# l8 P' ?
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递归思维要点1 a5 |; V8 ]( f' y/ c
1. **信任递归**:假设子问题已经解决,专注当前层逻辑1 A7 M/ \0 G! G& X$ i2 H
2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
7 J+ g/ Z/ x% h, W3 k2 r8 M7 \3. **递推过程**:不断向下分解问题(递)
5 h, [1 @# p, o  {' A( j4. **回溯过程**:组合子问题结果返回(归)/ n5 e5 c0 Z; ~/ a

  k' I$ h9 A2 ^8 w 注意事项& \- o# K, i$ R/ s6 C$ f5 b
必须要有终止条件& X% E+ l: D1 a9 P5 H3 o
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
% g& ^, ?: b: R某些问题用递归更直观(如树遍历),但效率可能不如迭代
4 r0 r, O+ C3 h9 p尾递归优化可以提升效率(但Python不支持)$ z% q9 a) i! R- `8 u  Y: o0 ^

6 f. I+ ^; M8 N: M" C- d 递归 vs 迭代
9 h( @/ i: k3 Q|          | 递归                          | 迭代               |% E7 b8 `$ u' F  _; X. F: L2 @! G
|----------|-----------------------------|------------------|8 C" R- O7 }" o1 s4 m
| 实现方式    | 函数自调用                        | 循环结构            |
5 Y6 B. r- |+ k5 u| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |: V& I4 g/ G3 U
| 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         |+ j3 L- g* f% X1 @! W/ o
| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |9 Q( s1 A4 B) c* @1 W

% O, I1 j8 j' [7 Z$ h 经典递归应用场景' X- k0 {0 ^, F2 V
1. 文件系统遍历(目录树结构)( }" j) ~: @- `( H8 U
2. 快速排序/归并排序算法' U1 i6 z% |1 w* {# w7 C
3. 汉诺塔问题
- E+ Q& n( A5 D6 i9 t) A. [4. 二叉树遍历(前序/中序/后序)
( k' f8 g; u8 A5. 生成所有可能的组合(回溯算法)6 F7 z: s/ }& I& F

4 z' S9 B: V5 o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,
: l* b' V% [( |- l我推理机的核心算法应该是二叉树遍历的变种。$ N$ h1 C5 o, }) }$ ^' y
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:
' V. Y6 l6 @( f: z+ @4 S5 {' F8 SKey Idea of Recursion
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A recursive function solves a problem by:1 W4 V5 }+ W! n6 e" T- a2 @
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    Breaking the problem into smaller instances of the same problem.
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  f/ c) t. v+ B. k    Solving the smallest instance directly (base case).5 Q% D7 A# b' t$ u* L
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    Combining the results of smaller instances to solve the larger problem.) A5 Z4 i6 O9 G3 D$ c
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Components of a Recursive Function3 o% {" n/ Q1 P+ q# k
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    Base Case:
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        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.7 y0 ]5 X$ O. V

' G  C5 v2 a( c        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.
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    Recursive Case:3 h% X; O( K1 r+ R4 Q

) B/ f( a9 @8 Z9 H        This is where the function calls itself with a smaller or simpler version of the problem.! h$ E% [2 w; i4 R
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        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).1 n' j2 \6 @; Q9 n' P* t

! d" ?3 U6 B! V4 u# {7 [5 RExample: Factorial Calculation  Y% J1 n3 g" V5 H, ^3 m; _6 J
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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:# l+ S( G/ x+ @: P) @
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    Base case: 0! = 1$ _6 r1 Y' Y9 k1 i% p$ Q; E

% v$ c' K& j. n! g+ f5 _    Recursive case: n! = n * (n-1)!
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& \8 U, Y/ j; Q! \8 q3 eHere’s how it looks in code (Python):
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def factorial(n):
. j" A- V" }8 x, V3 R/ S0 K    # Base case
. Y: y6 s% D8 e' r+ C( v    if n == 0:
! f( I' j/ X# V        return 1, G/ K2 e* R, n: v6 g- c
    # Recursive case0 ?/ l8 z9 ~! i3 N0 ?' S
    else:1 a% l6 }$ ?! }4 \7 W" ^7 x* G
        return n * factorial(n - 1)0 z/ T; B. f  v  u; I

0 U$ L8 h+ c8 b/ l8 V3 M  a+ `# Example usage
2 I7 [) E  ~% O& sprint(factorial(5))  # Output: 120/ c* o% `, ~8 c! D4 p) z
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How Recursion Works& e) [6 k9 f9 Y3 K9 x4 Q
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    The function keeps calling itself with smaller inputs until it reaches the base case.3 f% |' y* ?  _; i1 b2 w
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    Once the base case is reached, the function starts returning values back up the call stack.
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    These returned values are combined to produce the final result.: ]/ f$ j2 l; i

. \: L3 c2 E0 i& U8 h7 fFor factorial(5):
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3 d: r8 n" P1 v' {2 S* h  W2 efactorial(5) = 5 * factorial(4)
: V6 C1 S' n4 g$ C6 J# Yfactorial(4) = 4 * factorial(3)/ k6 ]& D( V1 r2 {- u
factorial(3) = 3 * factorial(2)
! k, C( ~/ N4 O  @7 ], o: c( a4 Hfactorial(2) = 2 * factorial(1)5 C5 h3 T6 Y, W, ?! I
factorial(1) = 1 * factorial(0)) g! r+ Y/ X& F$ K+ D
factorial(0) = 1  # Base case
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Then, the results are combined:
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# O, l; w* r5 K3 u% ^. wfactorial(1) = 1 * 1 = 1
6 ~- |/ z  |5 ]factorial(2) = 2 * 1 = 2
- F/ t* S) |5 @4 y8 K1 i3 k' L% }factorial(3) = 3 * 2 = 6
9 S( c! k* f( Ffactorial(4) = 4 * 6 = 248 @% W# A4 x8 e3 `6 ?
factorial(5) = 5 * 24 = 120& H8 l! H! p9 L' w
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Advantages of Recursion* h7 O" e8 a5 h

/ Q) \2 F/ m" J0 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).
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+ f. k9 c4 M& Z, ]1 A0 N2 r9 b1 a    Readability: Recursive code can be more readable and concise compared to iterative solutions.0 Y/ G6 K" i: }# D4 x
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Disadvantages of Recursion
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5 d' R. a! b& D& i    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.$ u, K; s: [' m7 t# e
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    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).) w" D4 j: |% Z# q& M& [

+ q0 Y1 M9 F" {; QWhen to Use Recursion
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    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).  w6 C  `" ^  w4 H+ @, Q
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    Problems with a clear base case and recursive case.
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Example: Fibonacci Sequence; g1 \/ c: ^, L

+ [$ N1 J9 [2 L  G' u4 zThe Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:7 P; R, j- r8 D) `8 N/ ^3 f

- L0 c2 X0 ]. X$ a. W    Base case: fib(0) = 0, fib(1) = 1
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    Recursive case: fib(n) = fib(n-1) + fib(n-2)
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python
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def fibonacci(n):2 B; F; J% g/ O$ k% _; I. N# E+ r
    # Base cases
( Q  H. z, T1 D, J; Z5 A# q    if n == 0:
" t2 u4 Q$ Q- R# L4 `" T1 i        return 07 o, [4 y4 L& K- Y+ h0 ~: h
    elif n == 1:
5 }9 R% X- Y. B; f! p7 N1 Q        return 1
( b$ a, c4 W+ X    # Recursive case
- ^$ r5 h, X* h$ o    else:8 q) m* H  y  w9 U- d
        return fibonacci(n - 1) + fibonacci(n - 2)5 @0 l9 n7 p# U5 q1 P
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# Example usage. U  m' Y& _% \2 l2 o3 i
print(fibonacci(6))  # Output: 82 Z+ h" ~1 v; j( ~- N; `- @
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Tail Recursion% C" T! b4 X# _  I: r1 y" Y" c3 ]9 G
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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).
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, M0 J$ _6 L) Q+ O# C6 nIn 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。




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