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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" W3 [/ m3 u; v: G5 {( q# @
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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。& Z* m1 N  r' D, _& u  r- Z7 H& r

) M5 O, B- [! q4 n" y+ Y3 S 关键要素
+ S3 b- h2 V5 {1. **基线条件(Base Case)**' u2 i9 O, g( @* O0 g- @" X% P4 C( o
   - 递归终止的条件,防止无限循环
5 h7 ~3 [$ X( P1 E   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1$ E2 K, E9 [- A  R
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2. **递归条件(Recursive Case)**! n5 Q( @5 W+ Q' r- O$ O# o( y
   - 将原问题分解为更小的子问题
% S: s& e; z/ v8 d0 V   - 例如:n! = n × (n-1)!
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经典示例:计算阶乘$ Q& i* \, b3 n0 Y9 G- y1 L' W& x
python: A% I; u( M; l- j4 X
def factorial(n):: i( w  X8 O. |+ h5 G0 U0 a$ n
    if n == 0:        # 基线条件
) E1 Z# `6 t* J; D' [        return 1. ^2 ]5 ^2 L  w; U7 V2 k! n
    else:             # 递归条件
! D7 A) j9 i2 C: S5 z        return n * factorial(n-1)
% U; z3 x- V7 A& e执行过程(以计算 3! 为例):. q2 M5 {1 G  [8 T  C
factorial(3)- s% ?! S3 w, i% d4 E
3 * factorial(2)
& T- @9 |+ C! T( E) ^& {3 * (2 * factorial(1))8 j3 M) f2 i7 K
3 * (2 * (1 * factorial(0)))3 ^' `! L0 c9 \3 l# @# ?" L
3 * (2 * (1 * 1)) = 6; \1 l9 Q+ g+ ]6 X

* h0 ~- i, \, o7 E3 g/ K 递归思维要点
) X! |' s% z5 r5 o  `8 F1. **信任递归**:假设子问题已经解决,专注当前层逻辑
( ]; R% W* T% J. p5 l2. **栈结构**:每次调用都会创建新的栈帧(内存空间)9 O! I" T' }& n1 R
3. **递推过程**:不断向下分解问题(递)' O5 q5 E' R! w
4. **回溯过程**:组合子问题结果返回(归)  p' ^5 X! \/ {( a( `4 @

- R# V$ t8 Z& j3 {5 w$ O; r 注意事项5 t6 w' b6 f/ {. T% n; f
必须要有终止条件
# }! }- R) t" p4 k9 l递归深度过大可能导致栈溢出(Python默认递归深度约1000层)- W& g$ K+ y; H3 s7 }
某些问题用递归更直观(如树遍历),但效率可能不如迭代
* M9 i) y: r$ p% n+ }& N7 z尾递归优化可以提升效率(但Python不支持). y# x# K+ \% N* y7 c
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递归 vs 迭代
- `+ |" [$ m. p! C, Y% W|          | 递归                          | 迭代               |
8 w; o) e+ \. i|----------|-----------------------------|------------------|
5 w$ q0 q4 Z- |) j' \- z| 实现方式    | 函数自调用                        | 循环结构            |
/ T; l- ~. s- R| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
: B2 f5 N" \. i3 o| 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         |' F" n6 K7 ^+ m
| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |8 G, c$ ^4 P7 A% o) s

) p1 i+ J) E7 ~: S 经典递归应用场景$ F% y' `$ r0 Z/ E" P4 x
1. 文件系统遍历(目录树结构)
2 C* y+ B2 @# F  S0 Y2. 快速排序/归并排序算法
/ P  s: m2 j# Q3. 汉诺塔问题+ K0 o: C8 Y3 n2 y
4. 二叉树遍历(前序/中序/后序)
2 z  i4 R0 ^/ H! w; ~& \5. 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,
8 }" o& c% V9 j! t5 p) g! I4 y) Q% W我推理机的核心算法应该是二叉树遍历的变种。( C  D: M5 s7 R. c* x( ~* H
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:" v0 N; \$ S' e) k7 H4 N# G
Key Idea of Recursion
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A recursive function solves a problem by:
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    Breaking the problem into smaller instances of the same problem.
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    Solving the smallest instance directly (base case).0 G/ m4 P5 f1 m7 E

1 G% m2 u1 h9 _( |    Combining the results of smaller instances to solve the larger problem.
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  _' D$ R2 Q0 _4 F) CComponents of a Recursive Function. Y% W" y4 j, J+ d; F

9 V& S$ H% d" Y    Base Case:( f  f; X; L* L/ y
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        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.' j, y) n" M/ s

2 n1 N9 ]  K: }9 Y        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:
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        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).- u" m9 T4 Z, V8 i# _9 D  |

0 D0 S/ B) w$ K) x2 ^Example: Factorial Calculation6 W. }. f& T5 q, M

5 `3 f" r6 G: B2 b; r* r! \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:/ [( W4 W. K" e+ X3 J! E2 p

4 Q) c5 R1 K1 |/ B' f    Base case: 0! = 1& {2 N/ g$ ]( ?# [$ `1 \. v

! z3 _6 K% _" I8 k9 a9 m  i' Z    Recursive case: n! = n * (n-1)!
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: i, \& F8 k& q5 n; d8 IHere’s how it looks in code (Python):
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def factorial(n):
/ w/ j/ U8 @7 z0 S( }, q1 Q    # Base case- U8 ~4 P% f/ J0 p
    if n == 0:
! f2 t* R0 y4 w        return 1# ?1 i) E* g* z' b9 a7 s0 J1 M
    # Recursive case2 u5 m6 f2 o* w$ H: g6 R
    else:6 b" [' d8 n3 g% L
        return n * factorial(n - 1), W/ j5 Y8 p8 n  `8 Z
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# Example usage, Z& Q( V/ s1 @9 U, j2 \& `( T
print(factorial(5))  # Output: 120
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How Recursion Works6 H8 ?4 s+ U2 [
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    The function keeps calling itself with smaller inputs until it reaches the base case.
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; w: L7 o5 t( g    Once the base case is reached, the function starts returning values back up the call stack.5 ]5 d9 R: i8 h
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    These returned values are combined to produce the final result.0 v! Y% [, p! m* y+ Y

  t: S; o7 \8 TFor factorial(5):. x/ Z, ]' s1 M) }' C4 Y
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/ U" }8 E% D/ C6 r' E% @factorial(5) = 5 * factorial(4)- I) i% Q7 R! Q
factorial(4) = 4 * factorial(3)  M8 `" e$ b$ \# u/ n
factorial(3) = 3 * factorial(2)9 i. o# u3 a& k7 |1 {4 H: m
factorial(2) = 2 * factorial(1)
0 X7 T& h% s4 U5 h2 i& [; I0 tfactorial(1) = 1 * factorial(0)7 ?; [8 K4 _4 Z% b' q
factorial(0) = 1  # Base case
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* {! h$ e0 H$ g6 g$ iThen, the results are combined:
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factorial(1) = 1 * 1 = 1
0 |7 A, ?- y1 gfactorial(2) = 2 * 1 = 2
3 [& R! n) K4 R( k, zfactorial(3) = 3 * 2 = 6* w. a' M3 S9 f
factorial(4) = 4 * 6 = 24# W& v$ x9 c$ z, U
factorial(5) = 5 * 24 = 120; |0 |% b0 O* C4 Z! Z5 s
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Advantages of Recursion
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+ E2 {. t5 ~+ d+ M6 ~    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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. T% ?( }3 {7 u    Readability: Recursive code can be more readable and concise compared to iterative solutions.$ q6 p3 ?( n5 t: w0 H! @

6 S; ~% u$ u& f8 b' VDisadvantages of Recursion
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2 B" M" S6 S6 }* o    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).9 K7 g5 R7 f8 d' k- U

5 ^7 X. g7 p* Q3 n9 K% OWhen to Use Recursion
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+ Q( y" H4 A2 i) \: W    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).( K7 B# u0 u9 V' S) f1 L$ q
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    Problems with a clear base case and recursive case.
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Example: Fibonacci Sequence! o' l/ \: u: I
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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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% l+ S, ?* V2 ~; Q& ?2 G% l    Base case: fib(0) = 0, fib(1) = 11 y9 W, E! k, B5 {/ x

4 X, M5 Q, U2 ]* m3 f( l' f2 P    Recursive case: fib(n) = fib(n-1) + fib(n-2)
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python3 O, D$ T& A7 G" Z

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* s7 y( o6 W! N0 R1 l" ?( ~def fibonacci(n):. }6 O- v3 z; @0 W2 y* j9 M6 n
    # Base cases7 G# |# Q2 e4 G- q! I+ i
    if n == 0:0 Y; i) O/ J' E# w: @  a
        return 0
+ `8 D0 z* X. m9 s4 |) I- p5 w    elif n == 1:
5 b8 U$ b" z, e7 G2 V        return 1
; v2 M7 C* a$ q4 b    # Recursive case4 D# f; ]8 y# V& r7 K9 a4 t
    else:
) t1 @/ S( I$ a- y/ X0 J" V. F, }        return fibonacci(n - 1) + fibonacci(n - 2). _, o- ^/ U' r1 c4 q

1 k- `+ \! N. k6 s. M8 e4 D! {2 Q# Example usage
  O; v) E6 {8 S2 cprint(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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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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