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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$ `: I0 U* J5 x8 F! [0 |7 ]7 [2 n
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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。
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关键要素) N0 R9 g8 z) J1 T% d
1. **基线条件(Base Case)**
+ r" o* s6 z6 i8 ^' [   - 递归终止的条件,防止无限循环
% C; M/ c5 J; j8 j0 d9 U. R* j, W   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1" z3 }  j+ T& O& q

4 P5 {$ m6 I, A3 L7 I6 \" ]& C2. **递归条件(Recursive Case)**! t9 x1 q% Q' g* i3 e$ y
   - 将原问题分解为更小的子问题
% e4 M9 g. u8 w6 d. u   - 例如:n! = n × (n-1)!+ o+ |2 m* g0 o

$ A1 \9 _$ D3 U- Y 经典示例:计算阶乘
* H3 V' J& K; gpython
& Z1 a, `+ D1 }& i9 ?def factorial(n):/ {3 O4 `, s. W
    if n == 0:        # 基线条件! c, n: ]3 A2 \+ @, y
        return 10 e' Q: |  ^' f5 ]9 N
    else:             # 递归条件; D2 Z( N) P* G- j6 x8 Q
        return n * factorial(n-1)
7 {' U' v, f, T; y- R3 w执行过程(以计算 3! 为例):
+ `: f' l4 k# Nfactorial(3)
" o1 C9 D; C7 `1 E- b3 * factorial(2)0 M2 k- e% M; O0 i% C4 K. n: r& u0 y
3 * (2 * factorial(1))
" M8 D- z. W+ ~2 u% N3 * (2 * (1 * factorial(0)))6 r# N2 ?+ g6 F- M
3 * (2 * (1 * 1)) = 6" n- m) M( T, g; E- h2 J: `1 I& E
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递归思维要点( G- y. ]5 w7 T  D2 n0 n
1. **信任递归**:假设子问题已经解决,专注当前层逻辑
. A) l! m" B) ^8 B* i2. **栈结构**:每次调用都会创建新的栈帧(内存空间). l0 T/ |' f) R. m, [
3. **递推过程**:不断向下分解问题(递), h1 v  |) j% M) J) a& ~) d
4. **回溯过程**:组合子问题结果返回(归)* \4 w) g3 t/ p4 ?8 S  k

) o: _+ I0 G- ?- F4 M  J2 V8 z 注意事项
1 c9 X8 {- ]+ A必须要有终止条件8 z; ]7 H$ M+ U( |' X' D: T' h
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
2 Y, ?1 O5 O) t# C6 ]6 W某些问题用递归更直观(如树遍历),但效率可能不如迭代
+ Q" f! L4 |+ C2 y9 H& |8 z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
; V' B6 {' j# Y% L) t, K|          | 递归                          | 迭代               |
- E$ n* j5 B( U. i8 F|----------|-----------------------------|------------------|3 Q, N$ |& F0 N+ ^6 K$ k* T! u* u
| 实现方式    | 函数自调用                        | 循环结构            |
! p5 Y3 b. K( L! S2 {$ u| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
) |4 w- v* W6 s/ ~| 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         |* D# y$ V  H$ b' X2 \
| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |' o; U2 g1 w6 u% U4 C7 d
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经典递归应用场景! f% l. H6 F& M7 {7 i# l
1. 文件系统遍历(目录树结构)
* W. B; ~* ^9 N+ e7 K/ f: l7 V8 \2. 快速排序/归并排序算法
" ^2 _1 t6 m0 e7 G1 I* Z/ |3. 汉诺塔问题
$ _% a" k) Y# i2 ?1 P4. 二叉树遍历(前序/中序/后序)
4 l7 N& M* y: v9 C/ ?8 K, ~& P' M5. 生成所有可能的组合(回溯算法)% G; o! K& Y8 ?, V3 v# q
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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,
9 R' x3 q: @2 e1 i! W我推理机的核心算法应该是二叉树遍历的变种。, p1 ~  B' P% c( k7 T5 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:" B' @, K9 S! A$ h. _
Key Idea of Recursion, P1 c" W$ Y. i' a' j
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A recursive function solves a problem by:, m  P5 d5 r' n+ u9 s6 {

5 @0 Y# [) _# H, @8 _# j1 ]    Breaking the problem into smaller instances of the same problem., u; I4 t! |# s* K4 }  ?

. s0 C+ R! N' u; I3 u& o. h    Solving the smallest instance directly (base case).
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    Combining the results of smaller instances to solve the larger problem.
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% S1 L+ G* x/ ^Components of a Recursive Function, k# `7 ?7 ^& H7 ^5 e
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    Base Case:
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, p  I0 j7 m: E( U! ]        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./ Y" v9 K+ o( I  N( n
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        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.9 I/ L4 V( A( q% g

9 w5 S( N+ N+ B9 Y2 w1 ^    Recursive Case:; I5 N! v4 ~- L& R

1 p+ Y, V3 m4 E        This is where the function calls itself with a smaller or simpler version of the problem.
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1 M0 P( A' I( m, W        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).
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. J, e, U! T: D7 h8 w8 ^Example: Factorial Calculation
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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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+ w3 z! L, ^4 v4 j9 k    Base case: 0! = 1' {! J8 I9 R) i, }
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    Recursive case: n! = n * (n-1)!, d' t" {7 t+ G  t8 @( b( S7 A, ~

+ p9 Z3 e# D- J- dHere’s how it looks in code (Python):5 V3 D1 v- ^1 x/ P
python9 ]: M9 ?7 L' J3 ~2 H; F% D
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7 c5 C1 U8 w8 k. odef factorial(n):
  @2 B; |7 s- X( k! A    # Base case0 N+ N/ w2 d9 ?. U
    if n == 0:
1 W1 o* W' z: Y& |% s        return 16 w4 L( {9 w, v3 U. ~. k4 O0 _. l# C
    # Recursive case0 X( c: r$ ]% L( `6 r
    else:) t' e$ g& }8 t$ h8 r* H( ]/ k
        return n * factorial(n - 1)
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; _2 d! m# V  Y8 _( A% p# Example usage
# t) a! Y1 X1 F. \/ pprint(factorial(5))  # Output: 120* `1 o4 J* {/ J' r: W# ~) Y# O
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How Recursion Works
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    The function keeps calling itself with smaller inputs until it reaches the base case.8 \2 e& W5 x  b6 O3 i
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    Once the base case is reached, the function starts returning values back up the call stack.
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2 E3 ]% [5 S; F3 P$ V8 c    These returned values are combined to produce the final result.
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For factorial(5):
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factorial(5) = 5 * factorial(4)# g4 Q7 u) d: g% t+ o! a
factorial(4) = 4 * factorial(3)- ~5 S, S* Z# v/ Q
factorial(3) = 3 * factorial(2)# @1 x- _# ~1 \$ p
factorial(2) = 2 * factorial(1)
4 O+ Q: @# v0 e  e2 Ifactorial(1) = 1 * factorial(0)( l2 Y+ C1 P6 v, V2 t
factorial(0) = 1  # Base case
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5 t0 e# L: d( b1 |( uThen, the results are combined:) n) d6 j: d8 W/ Y
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factorial(1) = 1 * 1 = 1
! z2 I, g: F1 {factorial(2) = 2 * 1 = 22 n0 |) m8 \& D8 h7 ~
factorial(3) = 3 * 2 = 6
+ Y$ M" u6 u1 P' vfactorial(4) = 4 * 6 = 24
+ E- y. z- J+ j' }7 q+ ]& i& [factorial(5) = 5 * 24 = 1208 m& V1 K7 p  d/ D6 C
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Advantages of Recursion
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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).9 Z- J2 R0 r; m! e1 Q- p1 V+ W

- r; \' j* v3 o/ k5 }( e    Readability: Recursive code can be more readable and concise compared to iterative solutions.
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( A8 D! J2 t! Z: c: a+ [) QDisadvantages of Recursion, ~8 [; `5 G  [5 d$ Y1 \! j

3 D( g$ R1 w' ?    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.* X% `% \/ Q$ B# O/ Z- J6 e4 s! I# k

* b- A! D8 C7 A    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
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When to Use Recursion* D4 A) X( z7 v% f$ F) X# k

2 {9 m6 p+ q( ~    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).3 r( |; P. B, W& O
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    Problems with a clear base case and recursive case.8 U: C3 T$ L% ^- _, C' P
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Example: Fibonacci Sequence
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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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    Base case: fib(0) = 0, fib(1) = 1
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3 i7 r( A: {3 _) B( z4 A    Recursive case: fib(n) = fib(n-1) + fib(n-2)
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2 B  u' ~3 \9 O5 ~; O% kpython8 F( T- W- Q0 m5 }

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def fibonacci(n):
* O5 ^+ _  r* i& {7 b  X    # Base cases6 ~& F. [; N" i$ g+ f0 R
    if n == 0:, i6 {% C/ @" w; T% f
        return 0
3 f1 q7 {3 `' O, F1 b7 e    elif n == 1:
' t3 @3 z! D' a        return 1
! k* l2 l7 L( t' h% m    # Recursive case
* U5 s% Y3 g* ?6 k4 ?% W    else:4 p9 S( W) A# X/ L) S6 {+ R" E
        return fibonacci(n - 1) + fibonacci(n - 2)8 V6 D! ~" K  U- y4 ^4 e

: r4 F2 G: P9 D( f! O1 A( K1 S# Example usage
, `# ]0 @& Y' f# F: [" P4 hprint(fibonacci(6))  # Output: 8. k' W! Z6 q. m
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Tail Recursion
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! T6 Y3 w. I  y* _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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