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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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0 I2 o" i8 w$ \# W8 [+ J解释的不错- E! V( i4 n0 H/ Y

! r7 |  V" k# s! b  ?5 @6 ^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。) c% T/ t$ q& a% g, j7 q* T
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关键要素
5 X: C+ h, E2 j/ f1. **基线条件(Base Case)**. Q. c, F) j, ?5 L. c5 G7 v
   - 递归终止的条件,防止无限循环
: {. b9 e$ x1 X) |   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1  D, r+ P& q" {% s& r6 X4 r
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2. **递归条件(Recursive Case)**
! v; Y) X* Z8 N4 V' T( T) w! [   - 将原问题分解为更小的子问题
7 B3 L1 C% @8 {* ^" h   - 例如:n! = n × (n-1)!8 k6 `6 a8 E% e( A
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经典示例:计算阶乘
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9 s# F/ F+ c5 u" a! A! e0 Rdef factorial(n):# e1 _, p4 o; o% `4 E% q2 F
    if n == 0:        # 基线条件. P' k$ ~" q: U/ u" ]: a
        return 1/ i" G5 M7 t' k# ~/ I4 W
    else:             # 递归条件( a/ v: n" G# ?8 W8 @5 Z
        return n * factorial(n-1)
: Z' @; M/ W' C9 J( _% I7 o  [执行过程(以计算 3! 为例):
2 R6 I! k- ^; zfactorial(3)0 w1 k1 G3 T0 Y5 R6 C! F1 @
3 * factorial(2)  r- p' k7 M. [  Y( T4 y! k
3 * (2 * factorial(1))6 i- c& }: q& b, Y% V  O
3 * (2 * (1 * factorial(0)))
- {8 w$ k6 Z9 B' R: i3 * (2 * (1 * 1)) = 6
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递归思维要点
( P. G0 l- }2 K0 T# o4 ^. x: i4 N7 z1. **信任递归**:假设子问题已经解决,专注当前层逻辑
$ t! b* H" y1 ~) N2. **栈结构**:每次调用都会创建新的栈帧(内存空间)8 }* [( X9 t3 e2 j6 O4 W
3. **递推过程**:不断向下分解问题(递)$ D# n% v% H7 \* _
4. **回溯过程**:组合子问题结果返回(归)& W6 T: ~9 P2 L9 c: g( T  C

- I+ O, I' G: |* ~# | 注意事项
7 [0 y# ]! O7 D必须要有终止条件" R6 e) ~! y) l
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)! p' U$ {; q) ?" o  V
某些问题用递归更直观(如树遍历),但效率可能不如迭代
1 A( M( Q8 c8 P- @  W3 e# C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3 L" a; r) x; E5 W- |
|          | 递归                          | 迭代               |( X1 d" g" x6 {1 R% k0 R
|----------|-----------------------------|------------------|
" p+ K9 g+ R- Z( j2 v& A  Q| 实现方式    | 函数自调用                        | 循环结构            |4 m; w: v7 ]" f3 @9 y$ H
| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
( a' m6 H' N, m3 x( T# _/ H% d| 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         |; h) a* \% v6 A4 Q: Z% P
| 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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. b5 P. Q1 q0 v) k( t# ` 经典递归应用场景
# Z. R$ m  d2 ^6 [% M1. 文件系统遍历(目录树结构)1 g  J% a- s% c" E7 ~/ k7 B
2. 快速排序/归并排序算法
1 Z' p$ `3 p$ ]$ u$ e3 X2 m3. 汉诺塔问题
1 ~( I5 _4 |6 c6 q2 ~+ ?3 R5 A$ K3 K4. 二叉树遍历(前序/中序/后序)
5 `% G+ E8 I  g: e1 x5. 生成所有可能的组合(回溯算法)
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2 \/ ~* `9 D% e+ z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,
6 ^( h. n& }! H. s我推理机的核心算法应该是二叉树遍历的变种。7 H/ F- i( e/ Z) I) a; x) }
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:  P2 N& P8 o/ o# a
Key Idea of Recursion" f, V" I) W; H1 i
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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.9 c% v" Y: H" v8 }/ V& t0 H  \; M

6 r. z/ q8 a* p  T+ O    Solving the smallest instance directly (base case).4 ^5 \5 W5 X% A9 f3 R

6 A5 _! a3 a/ ^7 H    Combining the results of smaller instances to solve the larger problem.
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( Q' J& G7 R6 g# HComponents of a Recursive Function
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( {0 T- a, |3 r5 A( K3 {3 W1 b. T    Base Case:1 g3 X2 p4 q6 @/ b
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        This is the simplest, smallest instance of the problem that can be solved directly without further recursion., D8 J6 {  j0 |1 {& _
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        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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% H  i7 a1 w' X1 o    Recursive Case:0 g1 O1 N& U: n' e  |

" d/ R+ G6 R# \6 E; \- \        This is where the function calls itself with a smaller or simpler version of the problem.$ t" }) S, W, N

% u$ H& t$ l7 d) _( M        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).
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Example: Factorial Calculation
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* N( s' e+ u% [1 h, H/ XThe 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:3 O* _5 m# b, e5 o/ F' A
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    Base case: 0! = 1
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, p8 m0 A$ {0 }9 U' c1 D' ^( X    Recursive case: n! = n * (n-1)!
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Here’s how it looks in code (Python):
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def factorial(n):
8 P$ p2 h1 b: Q    # Base case
( m3 @/ T$ Y' \& `    if n == 0:: S. m- t6 A' ^! X9 g* w. Y
        return 1
% S" `0 D6 L8 D2 Q! v, {( N# Q    # Recursive case- k" B$ O% C2 W
    else:+ B' ^# C4 j9 f* k" J# V# |
        return n * factorial(n - 1)
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8 }3 d/ w/ J6 ~) U: f# Example usage/ Q: W5 k* Q+ z8 f4 R
print(factorial(5))  # Output: 120, ^: c9 P( B9 j
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How Recursion Works
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* |. N+ c# _& E; A& L    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.  C, ^) S/ P- r3 c1 H, R1 v
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    These returned values are combined to produce the final result./ P; d& Z8 P) g/ _8 p4 I( U3 z9 o

4 g. v# C7 ~& o% k( cFor factorial(5):/ L# Z! l; U1 L
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factorial(5) = 5 * factorial(4). A; V; u/ j8 g* x4 {  V5 v
factorial(4) = 4 * factorial(3), O7 S  {' h4 d2 o/ x( u; R
factorial(3) = 3 * factorial(2)7 q' k1 D4 M- ?7 z: m! X
factorial(2) = 2 * factorial(1)
/ n2 U8 C! v0 \7 Dfactorial(1) = 1 * factorial(0)
9 w/ G! D3 J6 k3 k; Cfactorial(0) = 1  # Base case
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+ A9 q% O( o  D+ n2 \7 }2 NThen, the results are combined:
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factorial(1) = 1 * 1 = 1& \4 P+ O' d* V$ Z/ ]  g
factorial(2) = 2 * 1 = 2: U8 K5 R0 z! H. j' P! g
factorial(3) = 3 * 2 = 6
0 k2 x, {0 U2 `factorial(4) = 4 * 6 = 24; \& K- k0 V1 ]. U6 m! ~
factorial(5) = 5 * 24 = 1203 h. h" A' f6 J4 z4 V
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Advantages of Recursion  \8 W3 \& R! N* t0 _6 X+ s
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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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    Readability: Recursive code can be more readable and concise compared to iterative solutions.% Z  {% H+ S( C! G6 d7 x
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Disadvantages of Recursion
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, f1 @. M: i5 w$ y    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., o% M4 y& ?3 n' N
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    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).  p5 K7 p) ~: y1 f

- G; F1 n- W8 R" o7 nWhen to Use Recursion  q" @' U8 H- B- f; l6 P0 I

+ m8 b0 K' J5 h. q3 \1 X3 L/ @' U    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
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    Problems with a clear base case and recursive case.
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- o2 l4 I* {6 F4 V, aExample: Fibonacci Sequence& e! A* W6 Z. l8 ~  A: W6 N
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The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:1 J! E5 V. p: q3 A( z

' L* Y! }* f2 B0 K; f    Base case: fib(0) = 0, fib(1) = 1
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( g8 X# `$ ]) [1 g* v" R, G    Recursive case: fib(n) = fib(n-1) + fib(n-2)) D6 t  i" h- l8 o

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def fibonacci(n):
2 r/ e2 I0 t: t4 h) y! J& y6 G4 t2 g    # Base cases
6 G/ Z; q; e$ O! ^' e3 r    if n == 0:' S: Y/ A; P) H; O: m1 c# s
        return 07 ^" p) D- l* ^
    elif n == 1:
5 e  v/ x$ e7 `/ `* O) P3 o        return 1+ q# K0 G+ O  B4 l
    # Recursive case
. S. h# e5 C/ R( ?    else:
9 J3 H$ N5 B2 Q/ @6 z; S% W% W        return fibonacci(n - 1) + fibonacci(n - 2)
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0 @- m: D# Z& q. P, [4 S# Example usage9 ?: E  [/ d& m$ [1 W
print(fibonacci(6))  # Output: 8
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Tail Recursion
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* K3 t& F4 B3 P3 x5 K' ITail 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).7 s! u& T7 w5 ~/ J5 [
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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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