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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 2 F1 x# @; P2 h8 n) K$ Y# Y& q
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解释的不错* k4 K2 Q$ o0 [( y! A

" s/ e  l: A) E4 T# I  _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。
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+ v& W8 [3 _: o 关键要素3 |( Z! n7 _5 a+ ^* G& Y# N
1. **基线条件(Base Case)**. \( A$ p0 ]# d8 S7 x7 h7 z4 X
   - 递归终止的条件,防止无限循环
; `8 E& ?0 I& s- v% G   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1
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2. **递归条件(Recursive Case)**$ S! l: O( p: M( Q$ S
   - 将原问题分解为更小的子问题( E: i0 g, |7 ]/ F' o9 d* X
   - 例如:n! = n × (n-1)!
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& C. S7 d; r. T 经典示例:计算阶乘: `+ _  I$ E  ~: ?0 d1 j) c9 V
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def factorial(n):& v) L9 J* i& M% m1 h1 M
    if n == 0:        # 基线条件
5 `& ~9 a6 z+ a        return 1' P# l; V$ r8 I$ z' e
    else:             # 递归条件8 O, k$ w2 \) n, S; s
        return n * factorial(n-1)
  B* k$ A9 z/ N执行过程(以计算 3! 为例):7 a* f, X5 c, b/ z2 b; [
factorial(3)7 a$ `+ C2 j. t% L- G2 G$ W& J
3 * factorial(2)
- O' m8 P+ o* W9 Q9 a( _; t7 R3 * (2 * factorial(1))* Z) ?1 {8 c3 ?' i1 K" O
3 * (2 * (1 * factorial(0)))# }+ P. }0 i5 X
3 * (2 * (1 * 1)) = 6- ]1 K& S* @9 g/ Z2 d! j% }  h1 F
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递归思维要点0 D4 [. J# b/ L7 i; N
1. **信任递归**:假设子问题已经解决,专注当前层逻辑
1 c  I, S3 |* C4 Z6 C) ]# R2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
7 a. ~6 W, X  e! V- b0 C3. **递推过程**:不断向下分解问题(递)# _7 }6 b$ G* F0 j/ ]9 @
4. **回溯过程**:组合子问题结果返回(归)
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注意事项
: X2 T2 |; [0 z, _必须要有终止条件
. p% A2 y) Y( c  F, T. _% ^递归深度过大可能导致栈溢出(Python默认递归深度约1000层)3 J& E5 J: s2 I& J$ U% r# t% i
某些问题用递归更直观(如树遍历),但效率可能不如迭代
1 v* D! ~& U& s' f) b尾递归优化可以提升效率(但Python不支持)- B* W5 W8 j' }5 n# d& ^0 v4 h
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递归 vs 迭代
: f; S, G2 g3 \8 p4 w3 L|          | 递归                          | 迭代               |- D  T6 ?# m2 w
|----------|-----------------------------|------------------|$ o8 Z, z  U. S& K/ j  E& h
| 实现方式    | 函数自调用                        | 循环结构            |
1 N- X- {; _$ Q( H/ p! ~% _8 M| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |; J$ {+ ~9 n2 O4 i* \
| 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, v6 ~6 D, k7 \) U
| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |1 W+ c) ~. Q) K

1 i' S  g6 s; B- w" ?. f) s 经典递归应用场景& j) l! R5 _4 I6 n; H( O: b
1. 文件系统遍历(目录树结构)
8 r. ?2 d/ n- S5 x2. 快速排序/归并排序算法
3 N, O& ]+ Z- J3. 汉诺塔问题/ D0 j6 D* L; I. F! ]( u* `
4. 二叉树遍历(前序/中序/后序)
6 l: H4 h* n$ L# ~) k7 r8 K5. 生成所有可能的组合(回溯算法)
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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。
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,& t8 l5 x  T% p. z8 _. q8 [
我推理机的核心算法应该是二叉树遍历的变种。
$ z0 |6 f4 ]+ o0 J- e, @) s: x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:6 s) `! g, C) }: ^- {% P; Y$ O
Key Idea of Recursion, I% j( l' Y" z% {" h% F2 ]
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A recursive function solves a problem by:
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7 S- v4 e( P/ [: V: r    Breaking the problem into smaller instances of the same problem.
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0 M% }8 A# j8 A3 m) m    Solving the smallest instance directly (base case).& z5 c& U5 h& p

' r. S# q7 P, t6 _/ o    Combining the results of smaller instances to solve the larger problem.
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Components of a Recursive Function+ Q5 ~/ O1 y- M9 w% L
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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.) S% R5 v) o; u) p% \0 ~% ^' k

' _& j  }* K5 y9 E        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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    Recursive Case:/ e+ Q& z1 b- z; s0 p8 h3 K# \
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        This is where the function calls itself with a smaller or simpler version of the problem.
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; @; l8 }0 U" e- d        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).; w* c- ?8 j8 E0 H2 b7 n

9 K' L- t' p; _9 hExample: Factorial Calculation" i5 o* G7 m& U1 ^* n  `
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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:( x( T) m" E" ?* N& S
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    Base case: 0! = 1
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    Recursive case: n! = n * (n-1)!6 L, h8 ]" d! G! V5 O

. M6 d5 i5 g& e! O9 fHere’s how it looks in code (Python):0 _3 \5 m) W) ^* |" `8 o, K, U
python
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' x  n7 V' T; T9 s, l, pdef factorial(n):% A1 t3 O6 N" r  q9 E  ]
    # Base case
! ^( U4 @. p3 r- R/ m    if n == 0:, T9 ]" g8 l4 ?0 t) O+ `* [
        return 1' F- r, I; I+ I" l. |
    # Recursive case
; M1 Y6 S* R% u+ G! }, V6 J4 u    else:8 W- Z% ?+ }4 _( T1 X
        return n * factorial(n - 1)
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, Q( o" a) G; X2 U0 y2 [% w6 h# Example usage5 D. |  K  B4 {8 \
print(factorial(5))  # Output: 120
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How Recursion Works3 T( i' S4 k% s/ U' d  e8 g: L3 U" \
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    The function keeps calling itself with smaller inputs until it reaches the base case.
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* ]2 x# G# |1 D7 Z    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.
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* L  @' L. s( n) EFor factorial(5):7 h# j9 t; w8 m

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# k( }2 ]0 C% J4 tfactorial(5) = 5 * factorial(4)
( K* p7 O# j$ |; Zfactorial(4) = 4 * factorial(3)
# O$ F7 V& z4 ifactorial(3) = 3 * factorial(2)
/ T3 ~  R8 E, a" X$ dfactorial(2) = 2 * factorial(1)2 @5 B! A5 g' t4 |3 H8 N% e. @" e
factorial(1) = 1 * factorial(0)
+ \8 h; ~8 _9 M9 o  F  ?. E6 O) Wfactorial(0) = 1  # Base case% @, n! \) G' U/ g% I% m
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Then, the results are combined:
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factorial(1) = 1 * 1 = 1
; i0 \% M) F- I, Ofactorial(2) = 2 * 1 = 2
* i9 b6 I; A" [" i1 Pfactorial(3) = 3 * 2 = 6" ^1 U+ m* j/ X6 t# |+ L2 r
factorial(4) = 4 * 6 = 24- I7 C* D" `9 h" i0 }
factorial(5) = 5 * 24 = 120
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Advantages of Recursion& q0 i$ A6 |  z7 F0 {

; E& ^- O( f2 {/ b9 V. o# i    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 _5 u# X1 T. G2 L% {7 T: b
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    Readability: Recursive code can be more readable and concise compared to iterative solutions.
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Disadvantages of Recursion
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% z4 v' b: {9 I0 D8 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.
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5 n& ]4 g% L/ g    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
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When to Use Recursion& Y! k' s* O2 U
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    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
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    Problems with a clear base case and recursive case.3 [6 j5 }3 V1 ?; ^+ [
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Example: Fibonacci Sequence" z8 }  C" S' H/ A4 M
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The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
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6 n- l" K& ~/ [, e( H9 R' b% P; ^, U    Base case: fib(0) = 0, fib(1) = 1* a+ I) l+ L! c6 Y2 {: ?

6 c/ q3 N- k( [5 P: F    Recursive case: fib(n) = fib(n-1) + fib(n-2)0 T5 e5 B  x' _' ~

7 H$ J$ g' _1 a! Vpython
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; G. G" w4 G) j! ?/ v+ c( u5 e7 ]: Wdef fibonacci(n):2 y6 y* x8 {6 B8 O, U& M' F
    # Base cases7 F: m3 A/ h4 @3 J) a( g4 |  P
    if n == 0:
3 r7 ^' V9 w# |5 m        return 0( t) m6 V8 J& w) }) L
    elif n == 1:
5 a/ Q! _" @1 {- R        return 1( K" y6 W) Y& g0 Z  r
    # Recursive case* x- u5 ~7 d0 ~) Q+ E# ?) ~
    else:8 E0 ~6 `& }' z
        return fibonacci(n - 1) + fibonacci(n - 2)
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# Example usage, ~+ S4 s+ p  L$ k, p3 U8 s
print(fibonacci(6))  # Output: 8
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. }( G) ?, e- F2 W# y! lTail Recursion; J: W8 F4 }4 b/ m

: R$ f5 f# e: o- V9 l: i% p! vTail 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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: y* S& u- V1 F1 C8 s- {4 cIn 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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