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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 4 P9 i3 I8 B8 k7 N
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解释的不错
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) e# `+ v& Z& c* H/ a3 l& 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。
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7 |  K% O5 r& ^% C' i1 I 关键要素9 F! w. y7 Q: l) H+ d- r
1. **基线条件(Base Case)**! Y: j* [! w6 m! E
   - 递归终止的条件,防止无限循环3 e) K9 U6 |. p  `" x* }. k+ v
   - 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)**+ {9 c4 l$ R5 h
   - 将原问题分解为更小的子问题
0 b$ M1 j" v1 b. f   - 例如:n! = n × (n-1)!2 L% q9 Z% Y& }' R$ b

0 q- g8 S" Z( i, k 经典示例:计算阶乘! R# M4 g% M' [( z5 I+ }
python4 k' S- X- t! A) }+ c% r; G
def factorial(n):( M. o+ [! x: i, S4 z$ R
    if n == 0:        # 基线条件3 U! x4 q) v$ H* v
        return 1
5 K* M$ V- Y& \5 L( N# m8 ^, _9 |$ Y    else:             # 递归条件) S2 R, r8 s6 d/ m7 }
        return n * factorial(n-1)  P0 g; s$ p/ r% V
执行过程(以计算 3! 为例):
+ w: T; z3 @2 @1 b9 xfactorial(3)
2 t# }- z+ t, n# l6 W4 Z5 A3 Z: o3 * factorial(2)& @8 `4 \( L& `3 N4 F: Y" @6 X9 n
3 * (2 * factorial(1))
& J" j( i9 `3 e( R3 k3 * (2 * (1 * factorial(0)))
9 a1 u2 ~. b& V; ?% Z# M1 j, U3 * (2 * (1 * 1)) = 6
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, v9 z5 z+ C; Y- i 递归思维要点' ~" ?  [1 w' a! Y) U3 Y* t
1. **信任递归**:假设子问题已经解决,专注当前层逻辑
8 V* y# O4 H* s( ]2. **栈结构**:每次调用都会创建新的栈帧(内存空间)0 _$ {* Y0 X* j- _/ i
3. **递推过程**:不断向下分解问题(递)
3 T, q& l: T' v4. **回溯过程**:组合子问题结果返回(归)
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注意事项
2 v6 w, n# z0 b7 I& M9 p0 u: c必须要有终止条件
3 ~) I( x9 E. @递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
( Z: z8 X3 W# J+ h  m某些问题用递归更直观(如树遍历),但效率可能不如迭代$ S& R( ?' f9 C$ A
尾递归优化可以提升效率(但Python不支持). R) k) s. F( m2 t$ w" J

/ v, w8 z5 _) |4 N 递归 vs 迭代
; _+ a' {, r; z5 H+ z. e% v* T|          | 递归                          | 迭代               |
$ b6 Z* k/ R) k! T|----------|-----------------------------|------------------|
7 J& C& U9 z( m+ i. }9 }| 实现方式    | 函数自调用                        | 循环结构            |
  U0 I0 \9 Q  A: m| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
* y  r9 H- f! }8 j' m: |+ l| 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         |
- T- B5 c# f( x: U, [" [5 A6 U| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |. x' I0 s( S) _% T) i
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经典递归应用场景/ r6 n. T* o/ u8 B% `7 z
1. 文件系统遍历(目录树结构)
# O& y4 l0 O0 P# y" v. p2. 快速排序/归并排序算法
0 B2 }, F0 o5 Y3. 汉诺塔问题
2 H; T6 J5 X& S# }& H$ a; P4. 二叉树遍历(前序/中序/后序)
: C# r" O- h, H$ ?% x" U5. 生成所有可能的组合(回溯算法)8 N9 {" E, f- z' I& @. y9 @
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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,
2 k  ~! |( x3 o* B6 ?. i* s我推理机的核心算法应该是二叉树遍历的变种。8 ^6 G' n, l. ]- e
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:$ k8 ~5 ?4 d& W+ L, U; X: U$ N
Key Idea of Recursion8 n' m. S: M; `& ^- f9 ~" P

, {; x: M) v/ E+ ?  dA recursive function solves a problem by:- _! _8 }$ [8 x3 a

; H& @" a" y8 x% [% G; O2 C% y    Breaking the problem into smaller instances of the same problem.
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    Solving the smallest instance directly (base case).
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7 Y; r! C3 W$ H/ t+ L) y( Q    Combining the results of smaller instances to solve the larger problem.
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" ]$ ~' J0 Q; {0 QComponents 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.  m  J( q( J+ a+ m+ ^

5 [& B# i" ?# C        It acts as the stopping condition to prevent infinite recursion.
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: O' f  L- ?9 i8 c3 |! T$ K        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.% \( }- _- A$ U$ w/ k; j, T/ k

$ R1 d& S& P2 \0 V    Recursive Case:
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7 i3 s. T( ~$ g        This is where the function calls itself with a smaller or simpler version of the problem.
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        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).% g4 D6 o5 i. e1 m) H2 I8 }

' V' S; ~( N8 D' H' V+ g' u2 D7 M) }Example: Factorial Calculation- f7 G+ b+ {' J: G- D+ O2 Z# ]& O8 A

6 z) |- u0 i+ x* R* y) ~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: S; q4 F5 y* Y- X( N3 S" f
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    Recursive case: n! = n * (n-1)!
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Here’s how it looks in code (Python):+ L) s1 s. e3 [) f
python
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9 N& p3 }1 f4 {7 Odef factorial(n):
% c. w# @3 N% Q; G    # Base case7 n# g7 ?/ `# Z. c1 c2 Z3 M
    if n == 0:+ W  N8 l& M  U9 X$ n& l
        return 1
4 H* r) G) X) o3 j% H# m- D& c7 t; |    # Recursive case
/ b& n* }+ R+ g9 c    else:, t' H$ u8 k$ m+ B- t, M: b
        return n * factorial(n - 1)8 L" c5 b0 s/ {6 V/ P- \: O
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# Example usage
4 }  ~" |2 r' Nprint(factorial(5))  # Output: 120
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How Recursion Works+ C. E- S5 \9 V$ r

( D6 |& B" }9 l* I3 ]1 I& _: _1 y    The function keeps calling itself with smaller inputs until it reaches the base case.
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    Once the base case is reached, the function starts returning values back up the call stack./ {; a7 \6 D) O6 t; N0 {

/ n! S; o% ?& ?+ o5 ^- x  p1 `    These returned values are combined to produce the final result.
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) T6 h$ I6 N' KFor factorial(5):
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9 Q  T( K9 z1 gfactorial(5) = 5 * factorial(4)
- l; F- a" m# |( v9 Zfactorial(4) = 4 * factorial(3)) M, b, _6 G# M$ t6 @  c2 I& v7 p; o
factorial(3) = 3 * factorial(2)( s% V- o" h% {3 H, l0 G
factorial(2) = 2 * factorial(1)
0 w  F$ g, Y5 Ufactorial(1) = 1 * factorial(0)5 X* `! {  L& Q; j
factorial(0) = 1  # Base case
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Then, the results are combined:  \/ d+ m/ X! h' ?

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3 z5 M  [+ E* hfactorial(1) = 1 * 1 = 1
( J, Q3 |0 M) t" h3 d* Y6 \factorial(2) = 2 * 1 = 2
- ^* V: B8 D- n1 O  e7 }factorial(3) = 3 * 2 = 64 ]# m; c" X# [, D9 f/ T
factorial(4) = 4 * 6 = 243 [% F. V& [4 }8 U8 Z# q
factorial(5) = 5 * 24 = 120
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4 K) o/ w/ @7 f4 R- Y& oAdvantages of Recursion  \( d8 z3 v- d5 k" p( T
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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).
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/ t2 l6 b; L% G! Q    Readability: Recursive code can be more readable and concise compared to iterative solutions.
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$ C& ~" }- q" t9 v3 H8 ~: F% Y* ?. @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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    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
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When 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).- Z4 R" C3 _, _( O
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    Problems with a clear base case and recursive case." R% l  N* L6 R0 D3 Z
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Example: Fibonacci Sequence
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- G/ Q' b8 ~9 ]4 q8 DThe Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:  @* e$ [( L) x# k6 L

& D/ T, ~8 w6 Y+ {! i    Base case: fib(0) = 0, fib(1) = 1+ W/ X! _( ?$ O/ D. `

- o8 e* M0 i7 I" V2 f1 [    Recursive case: fib(n) = fib(n-1) + fib(n-2). W* E- @1 ?& {  R9 {( n
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python
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def fibonacci(n):
  e% b: R) P, T7 i+ r2 L5 T    # Base cases& p8 P4 N0 n8 |& U3 h2 u
    if n == 0:' z! @6 H/ |: Y$ L: X) W
        return 0
3 i. f! d' C+ D) O    elif n == 1:
9 y- n" [7 H4 N' R0 B9 ^        return 1
) z* N" W. P8 y    # Recursive case- D4 p; \' x2 V, T; ~$ c
    else:
+ l, K/ |! X( y/ n3 T1 b        return fibonacci(n - 1) + fibonacci(n - 2)
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# Example usage9 e) B7 l3 b3 H( Q  R; C! b2 P& R
print(fibonacci(6))  # Output: 8
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Tail Recursion8 r; @8 x( U. t, T- @- u

, Y. c8 t8 Q6 ]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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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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