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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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# S. l/ u0 o* W5 R: c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。% K" a  D: L; ?% g% a
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关键要素
. z$ o( m+ N, ^9 N. n5 r+ q1. **基线条件(Base Case)**& t/ ]; W% u& n5 y0 Z; l
   - 递归终止的条件,防止无限循环
" D1 D0 X$ l# n1 \5 C6 X   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1* ^- z3 f/ r! K3 l# p3 n7 k5 h

  C; U1 p1 [$ ?/ K! r3 Q% k2. **递归条件(Recursive Case)**
: l  C& `* V+ v4 }/ W" ?   - 将原问题分解为更小的子问题
) d4 r! {( a, J. B# D. g+ _   - 例如:n! = n × (n-1)!
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经典示例:计算阶乘
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/ g5 w& o$ N! odef factorial(n):9 I' l+ |" D3 |) O
    if n == 0:        # 基线条件; Z+ y6 u8 {7 e9 G: Y7 @8 D+ f
        return 1
- z# o3 l% O- b) I1 x. n    else:             # 递归条件
, ^) g9 f( ?: m, G; M5 `  ^        return n * factorial(n-1)# Z- d3 f; o( R) ?% X: |
执行过程(以计算 3! 为例):2 H5 V; u" i, J- Q& _
factorial(3)/ h3 j8 _# T. E/ z2 C3 T* k' k
3 * factorial(2)- N) z- |$ }# S( c  W6 u. d; ?; w3 r
3 * (2 * factorial(1))
' j9 v5 E! x4 C8 x3 * (2 * (1 * factorial(0)))
$ \( M  O, x' {3 * (2 * (1 * 1)) = 6
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/ f# \% U3 ?' k  _! F 递归思维要点* A- }3 h, K* j" U: C5 Z
1. **信任递归**:假设子问题已经解决,专注当前层逻辑; k+ M0 ?0 f2 r
2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
' j# k7 C+ B; s3. **递推过程**:不断向下分解问题(递)
1 ?0 t, Y& N" {3 s, t+ a4. **回溯过程**:组合子问题结果返回(归)  v3 X4 H: Y- ?: K. E. \  M0 a
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注意事项
  u. a+ N0 g" V7 r% h- c必须要有终止条件8 p& F2 m, `& F7 E# M( G/ C4 H
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)) D' c& W/ E4 B: k
某些问题用递归更直观(如树遍历),但效率可能不如迭代
8 e7 T& t2 u, ^/ k6 Z" V. b" L尾递归优化可以提升效率(但Python不支持)
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+ \/ Z  D3 ?+ ^" K5 L 递归 vs 迭代1 K/ W" K6 y; Y0 A  H* s3 y
|          | 递归                          | 迭代               |6 o4 X. C! S* a4 U( R8 E
|----------|-----------------------------|------------------|
, V  w' ^" [  z2 n7 U: W| 实现方式    | 函数自调用                        | 循环结构            |
( g: ]5 Q$ n: Z1 m0 r* T" l| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
8 L, S: \0 A4 j  Y4 U% A  L7 \| 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         |* a6 W+ F$ Z' ~% U3 W
| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |
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经典递归应用场景6 A" w  t1 {& W
1. 文件系统遍历(目录树结构)! s& a6 v; U% ^4 e% Q
2. 快速排序/归并排序算法( c. x4 G! j4 l
3. 汉诺塔问题. ?  L, ?; i8 Y( A7 G* z
4. 二叉树遍历(前序/中序/后序)
9 `- e1 ]) E! D% S& q: n) {3 ^5. 生成所有可能的组合(回溯算法)
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1 h: t% e) v: k! I7 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,
- S9 r5 X3 R9 ~  L1 V我推理机的核心算法应该是二叉树遍历的变种。
: [4 P- }& y% R3 y; E) f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:
/ |3 E* v5 X. S6 b; D4 F0 }# ?' cKey Idea of Recursion) l' q3 A4 S2 s1 W- d

$ D' k- V6 R  B7 bA recursive function solves a problem by:
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    Breaking the problem into smaller instances of the same problem.# I0 a2 c1 S8 @2 W% \% O3 c

1 k+ T6 A* l+ ^) y    Solving the smallest instance directly (base case).% R. D3 |$ ~# z% M, y( B
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    Combining the results of smaller instances to solve the larger problem.
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: F- K6 c* j9 _Components of a Recursive Function
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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.) X: d* U% N: D0 x, `; l* N% [

. f' F/ i. W, |        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.
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& ?. J/ J3 l% p& l" L: v    Recursive Case:+ t8 ?, ?+ \8 x

0 ?+ E- R! C# H; {) P; U6 q        This is where the function calls itself with a smaller or simpler version of the problem.
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; X1 k! s/ f& T9 Z        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).% ]6 h3 z, L3 _; ?( _7 X) }

( H( f1 ^" J- YExample: Factorial Calculation- O3 H* E. w: Y" l6 M( \: D) b

9 ~2 u2 I( g, }" _7 r) d9 b- ]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:# u: S$ V% a+ d' @+ `  Y

1 V$ ?7 h. @& a8 g    Base case: 0! = 1
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    Recursive case: n! = n * (n-1)!
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7 ?8 U: X" O) h0 nHere’s how it looks in code (Python):4 ?( p# x) o9 U
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6 q- Q' _) \% t. Odef factorial(n):$ O% ~  [! {) A4 W2 P5 I2 c, m3 L3 ?
    # Base case7 U/ _4 G2 e3 d9 q; h
    if n == 0:
$ i$ w3 r1 [7 g8 i: G( @        return 1  x. y! M. `! ]( M6 `6 O
    # Recursive case
% z( T' u* R" {! P9 v7 y7 @6 }" t    else:
; |' ~* b: v% ?% a* \        return n * factorial(n - 1)" A2 B# ?/ g+ T3 K( Y

" I  X6 T' x& h! E# Example usage: E9 [. |3 t1 I4 D2 t
print(factorial(5))  # Output: 120
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How Recursion Works9 @0 q7 v- w* Q# l. m1 `. Y

$ N) u6 L9 i# G- @0 j' y) Z    The function keeps calling itself with smaller inputs until it reaches the base case.
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& d* Z* q- b$ V. B    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.7 s: C4 @) P: r
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For factorial(5):
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- ]' r$ ?$ f" b" sfactorial(5) = 5 * factorial(4)
1 D* G* A3 i- o; {6 }: h0 ^factorial(4) = 4 * factorial(3)9 Q# s0 R% M+ V
factorial(3) = 3 * factorial(2)
$ C3 O. ?- p, p5 ]9 r+ jfactorial(2) = 2 * factorial(1)- L. L- f5 C( f8 I
factorial(1) = 1 * factorial(0)  w+ W9 K$ x6 J% p
factorial(0) = 1  # Base case. D/ s5 n8 U. q4 |, ^8 ]

# n( u0 o- B4 i! E' j$ ~1 }5 z6 X" Q0 JThen, the results are combined:
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2 u* |5 S4 Q  C/ u$ U  ?* rfactorial(1) = 1 * 1 = 1- B. m4 |3 _0 R+ v9 i
factorial(2) = 2 * 1 = 29 v2 i3 u6 A" s' t) R
factorial(3) = 3 * 2 = 6
+ i6 I3 Q6 }8 F2 ]factorial(4) = 4 * 6 = 241 s3 i3 G( s! o
factorial(5) = 5 * 24 = 120/ e. K4 o$ O5 k2 w: W. J
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Advantages of Recursion
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) W6 h9 I1 E9 ]$ N- C/ c. [    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).$ x  C* L1 T) l7 y0 l, c1 |

  y6 K" P3 r3 s    Readability: Recursive code can be more readable and concise compared to iterative solutions.
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Disadvantages of Recursion
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    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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( T' B. M- ~* G; r# S/ w    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
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. Q. Q. \+ v! S; @( [2 WWhen 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).# a- S5 |7 l6 F7 o( p$ G1 ^

1 n  L6 }0 P* k& _/ t* \$ ]    Problems with a clear base case and recursive case.
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  o+ I5 S/ M: |0 c' |# S+ BExample: Fibonacci Sequence4 N+ e* z. m) [. A9 h( ~

1 d7 a7 Z- [# C, {The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:! k2 P' I6 A6 D0 f' p& ^- {& i8 [

* T' V+ H* B2 r% L+ E    Base case: fib(0) = 0, fib(1) = 1) A4 U* ?  y) ^3 T  q4 H

" H3 \: O+ ?3 O! d    Recursive case: fib(n) = fib(n-1) + fib(n-2), y) s9 f) a) H3 Y6 [
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python
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) m$ X& l' Q8 `7 v" ydef fibonacci(n):
% Y/ f& b# Z" @1 ^    # Base cases
9 K" O( f4 H2 F/ \* r    if n == 0:! m2 w% \+ F4 J! X$ a  c
        return 0
) P% _! [8 i$ Y  p1 ~& G+ l    elif n == 1:* t& R/ r  j5 W, \& x
        return 1; ?2 O& |( S& _" B
    # Recursive case) u- |6 {3 L8 M2 N
    else:
& c  w+ g: ^  P0 U; H' G        return fibonacci(n - 1) + fibonacci(n - 2)
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# Example usage
: r. u/ q0 J3 a& W$ K# k: V4 X$ Lprint(fibonacci(6))  # Output: 8
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Tail Recursion
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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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+ s. |5 q3 `: Y4 n' j" t9 N# hIn 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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