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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 8 A  B. p% ]  ?& m0 Q
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解释的不错
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, G* p7 }6 d/ m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。) t( m9 y4 `! w3 w, f8 ^
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关键要素
9 c) Z1 k3 w  g. n6 \; X1. **基线条件(Base Case)**1 Z( T. D6 c" B% Y, `7 k
   - 递归终止的条件,防止无限循环
5 s: j! ~( k$ x   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 17 v8 y9 F, w4 j" W

% y+ V' O3 z9 E2. **递归条件(Recursive Case)**
% d* o' p" Z( D) H% I" c   - 将原问题分解为更小的子问题% O" P; \2 t( i0 k4 m. z1 k
   - 例如:n! = n × (n-1)!
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经典示例:计算阶乘, d; P: m: E: p7 s% e+ W1 v' E
python
2 ?- A4 p1 ?/ l5 L, z* Qdef factorial(n):
0 D! Y8 a$ A* N    if n == 0:        # 基线条件
' L/ O( B! d! B. c3 j1 v. U  h        return 1
' P: g2 r+ m  y; J4 f0 U    else:             # 递归条件
' R( U! B  ~  u) G0 D: c        return n * factorial(n-1)+ s# H. q4 E: z5 F
执行过程(以计算 3! 为例):: P0 a6 b% ?. S1 }3 E
factorial(3)2 Z4 E" ^' N; a$ {# J. ?
3 * factorial(2)
9 I0 P1 q( A5 N3 * (2 * factorial(1))7 k$ J' Y2 C" X: X
3 * (2 * (1 * factorial(0)))4 Y/ A% X. u5 p  U
3 * (2 * (1 * 1)) = 6$ G1 Q6 T( o: [, s( z

3 I4 b5 `# I. t  z 递归思维要点; P; m3 O; W+ K4 X! t# z" t
1. **信任递归**:假设子问题已经解决,专注当前层逻辑
5 s8 e. v6 G3 x$ H+ P1 m2. **栈结构**:每次调用都会创建新的栈帧(内存空间)2 g3 [( @! L$ c6 {5 v
3. **递推过程**:不断向下分解问题(递)
: J# g& T( d5 U" R' J9 y6 A4. **回溯过程**:组合子问题结果返回(归)& D3 J2 v. `: a5 a0 K( [

/ J1 K9 }  \; H8 a. W4 R) f0 n+ T 注意事项
1 `9 K6 Y" z: n; {3 j3 v必须要有终止条件
- k0 O' {7 V. I+ o4 \递归深度过大可能导致栈溢出(Python默认递归深度约1000层)! y7 L; w1 W/ p- j2 a) i' L
某些问题用递归更直观(如树遍历),但效率可能不如迭代" O( a! c, l6 {8 H: M$ d
尾递归优化可以提升效率(但Python不支持)
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递归 vs 迭代# [# [5 \* U* E! l* `
|          | 递归                          | 迭代               |
* u" u$ E3 h! D# W" s; i* `|----------|-----------------------------|------------------|
- o7 }  J2 l0 C, }| 实现方式    | 函数自调用                        | 循环结构            |; _" v( o, `' A, w; m& |
| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
  d. A4 K1 W+ o, E+ `) x" Z| 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         |
4 {1 a$ v7 [: v0 ]| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |$ O* r0 y+ L# w8 E; v% r

- \4 F+ }1 S9 ~4 ^+ n8 d4 `' E 经典递归应用场景: {: m- [& \% d& V% M0 z0 S
1. 文件系统遍历(目录树结构)
, O4 l- L5 h1 B' g6 b2. 快速排序/归并排序算法
0 f8 o, O; S& j4 ^, {8 D/ n3. 汉诺塔问题8 w8 A: n7 J! P" x2 i; B1 N8 Y
4. 二叉树遍历(前序/中序/后序)
. f6 Y: ~: V0 w0 o5. 生成所有可能的组合(回溯算法)' x& ]9 ?2 r$ o; |  l3 T

0 U* w, U+ J( w# [" s! k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,
8 F# T/ H# L0 c7 l我推理机的核心算法应该是二叉树遍历的变种。
; J4 w  r+ g+ Z) b6 J& j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:! w, w/ y/ R2 k+ b0 s
Key Idea of Recursion
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1 ~, ]# O; u3 d9 I9 K; mA recursive function solves a problem by:1 Q& o2 W9 z2 W9 \

- C. [: [6 R' L" C* t& s" H    Breaking the problem into smaller instances of the same problem.
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' J  ~- i( j# l! ?% ~) y    Solving the smallest instance directly (base case).! n  F! _" L$ t
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    Combining the results of smaller instances to solve the larger problem.( \1 A! d- {: H# ?
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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.1 V; o8 P; [/ a$ ~$ I
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        It acts as the stopping condition to prevent infinite recursion.
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- Q/ x/ a4 ]3 Y8 D( m. }        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.- X8 o/ @- m% `; q3 ]. A
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    Recursive Case:
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        This is where the function calls itself with a smaller or simpler version of the problem.
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$ h: w, p! B- a  G' m3 D/ ?& Z6 j% W; G        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).- l& [& A7 B- i( p; |

! A6 B2 D/ {  ]8 v/ VExample: Factorial Calculation) x2 a6 v- t7 k( a5 A
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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:
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    Base case: 0! = 1
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! \+ c: i" _2 Y2 a- o& r6 M" A    Recursive case: n! = n * (n-1)!2 G& _$ ?" ?* s/ Z# b; a1 }5 K

; Q/ b7 y( j& C/ v& NHere’s how it looks in code (Python):; w1 Z: q! g+ d" m  Y" {  u: ]
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def factorial(n):
9 ^* p( |( I' L! n- `    # Base case
! n1 G% e4 A6 k: ~$ s    if n == 0:
. p4 ]) s3 Z  X8 M        return 1
4 N0 f3 h3 Z# k+ W) A" r    # Recursive case% m/ k5 B, B0 T
    else:
& f( \& t4 V# a        return n * factorial(n - 1)
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# Example usage
- J: y+ J4 J8 l, F! iprint(factorial(5))  # Output: 120- T9 o: q( J; Y6 g
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How Recursion Works
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+ h# O$ Q5 K5 x; L1 [* k/ k% d    The function keeps calling itself with smaller inputs until it reaches the base case.
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, W4 O' R" C2 h8 i2 [    Once the base case is reached, the function starts returning values back up the call stack.
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) d' |& g: Q$ d  {% `- Y    These returned values are combined to produce the final result.
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; p) U- y0 ?& G1 xFor factorial(5):$ m0 l) p; A1 T4 D+ u4 K

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factorial(5) = 5 * factorial(4)! M3 {  s8 u$ B4 ~* f3 n* i
factorial(4) = 4 * factorial(3)
( C. [6 |( L( A5 d3 H4 Y( xfactorial(3) = 3 * factorial(2)
% C8 q0 j$ G, R5 \* f9 E. Efactorial(2) = 2 * factorial(1)$ p. q! z7 X! n, N: f( A
factorial(1) = 1 * factorial(0)
6 }+ k! x; P- R# _# y2 afactorial(0) = 1  # Base case
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: j& y2 y; M& ~  MThen, the results are combined:
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: q. s7 N0 _+ d4 @  w, Mfactorial(1) = 1 * 1 = 1
: j/ A& n7 S/ q& Kfactorial(2) = 2 * 1 = 2) Y' o* b, K' m* d% a, ~+ `
factorial(3) = 3 * 2 = 6
8 `/ N$ I3 ]% U1 Zfactorial(4) = 4 * 6 = 24- n1 X/ e5 N' v
factorial(5) = 5 * 24 = 1205 I& e/ J7 X& [, l, x7 X

& ?( l& {( t/ k' y% b  l. oAdvantages of Recursion6 a. g0 B9 e# k( _* ]0 i
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    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).$ h7 H8 d+ @" X! o1 h7 o) T2 U
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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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" v# f5 k% N" B    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.1 |3 m4 H0 m. J3 |
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    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
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  |2 o, R" {& p& b$ u9 kWhen to Use Recursion1 [1 G1 S( Z' T6 M
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    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).. ?8 u+ g7 Y5 ]! k

) F+ k: r4 {! k    Problems with a clear base case and recursive case.
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Example: Fibonacci Sequence! [" j- M* P1 e% D

- T7 T6 B+ [0 p+ L" |The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
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    Base case: fib(0) = 0, fib(1) = 1% R# K# C( {. k- }5 {; O

' {  |4 k( J5 f3 Z, x% r    Recursive case: fib(n) = fib(n-1) + fib(n-2)' O+ S& p; `! T, G1 X+ F+ `4 Z' D5 ~
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& b9 u8 B  z5 r6 N  r! jdef fibonacci(n):
7 J, z* b9 Z+ S3 R" @  K    # Base cases1 j" d3 z" E# A3 E7 j# S
    if n == 0:! [& U* K( ?* Y3 r& C0 M+ i
        return 0+ ~, x7 Z/ u6 T0 @& n
    elif n == 1:
& ?/ X, q6 _8 [- q        return 1
9 U1 R) b- A- k4 n8 d  e    # Recursive case
, O4 ^& e( Z# G; z1 e    else:
0 w% ]( J. ]0 H7 f* ?4 |) ^        return fibonacci(n - 1) + fibonacci(n - 2)2 g5 M3 {+ q: J# E* D9 i

# z- N5 w8 k+ z6 [* V0 p# Example usage
& x+ f# n, a- e, R% Oprint(fibonacci(6))  # Output: 8- p9 Z' G  [9 A8 n
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Tail Recursion
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9 d6 w: R; X, Y3 i8 o) CTail 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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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.
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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