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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$ I- Q! e  p# R) ^' z  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。
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关键要素
2 E' s% b1 N+ u! A# y/ O: m: e1. **基线条件(Base Case)**( e* o: t# W0 ^+ u* A
   - 递归终止的条件,防止无限循环
' C& Y3 M8 Y7 s   - 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)**4 f0 X# ^: \3 u! N( `& L
   - 将原问题分解为更小的子问题
, F; o1 P" i7 t4 U1 o   - 例如:n! = n × (n-1)!8 j8 G5 g/ ~9 H  h$ m" D1 k
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经典示例:计算阶乘
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def factorial(n):+ j- r4 E7 N8 |& k5 f6 r
    if n == 0:        # 基线条件
& G# S, v( V$ K2 G) J: n' j. Q0 H        return 1
8 L5 g, E  F% e8 N    else:             # 递归条件
5 z( W0 X- q) e0 H& [3 l! b        return n * factorial(n-1)
2 ^/ x7 T3 T! q3 K6 W. L% L执行过程(以计算 3! 为例):# {' [- q& p. W# y0 _; j
factorial(3)
. c+ ?1 w6 i* h# U( o3 * factorial(2)
# J0 w+ B/ u6 g2 W' s5 m3 * (2 * factorial(1))/ f# t  ^  G  k$ S1 L( M3 ^
3 * (2 * (1 * factorial(0)))* d/ P* H9 F- u, k4 O6 g
3 * (2 * (1 * 1)) = 6
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" N1 g6 m- K6 K/ Z( y 递归思维要点0 o$ G) L% N1 L: ~& V( o
1. **信任递归**:假设子问题已经解决,专注当前层逻辑9 j# j+ s  J" J, P( @
2. **栈结构**:每次调用都会创建新的栈帧(内存空间)9 U5 b1 P9 v$ O) i4 E% P
3. **递推过程**:不断向下分解问题(递)4 `# t, l' [, M. e% D
4. **回溯过程**:组合子问题结果返回(归)
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注意事项7 y, q1 G8 ?6 O. f9 w
必须要有终止条件: m; P! w" |9 d# _
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)+ B9 ^- s% A: I2 @9 U: h. Y% B% \
某些问题用递归更直观(如树遍历),但效率可能不如迭代
1 T6 a/ @+ W$ S3 n3 q2 i, l2 M- G3 l尾递归优化可以提升效率(但Python不支持). A$ a+ D+ M1 G) E4 x: A
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递归 vs 迭代9 M+ z. e8 K$ [% Y/ c# \
|          | 递归                          | 迭代               |8 t6 O% K  Q7 M9 v2 C5 x: F) Q
|----------|-----------------------------|------------------|; p+ R$ }) ?$ e3 N. H5 q, k
| 实现方式    | 函数自调用                        | 循环结构            |
+ ]9 i+ l( W' B. g6 c. P; b| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
5 t. a6 z' d; x: T1 u9 w2 w| 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         |' O' x! l0 E0 q6 N; a9 Y+ e
| 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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经典递归应用场景
3 N  H$ b  N* L- z+ v& ]! ^/ }1. 文件系统遍历(目录树结构)
0 f1 s& s* d' u; l5 A8 K3 }2. 快速排序/归并排序算法) i* H+ _% v5 Y3 u, ], O+ c
3. 汉诺塔问题* ^4 X0 w' Q$ I! g& s
4. 二叉树遍历(前序/中序/后序)9 d+ O, d" V6 M' ^9 T
5. 生成所有可能的组合(回溯算法)) G6 L9 ]/ v  F0 a( D4 S2 c
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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,
5 S0 ]4 g3 p0 c$ }2 p* [* L我推理机的核心算法应该是二叉树遍历的变种。, p* @4 D$ T  [
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:
( M4 q$ }$ y: a$ u% pKey Idea of Recursion
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A recursive function solves a problem by:' b' ?5 [$ m& ^4 S! _
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    Breaking the problem into smaller instances of the same problem., J  `0 v- R& q9 Q( x; n

, H5 d1 \  Q/ S% |  e# ^    Solving the smallest instance directly (base case)., S, G- h6 V8 Z

5 h% I5 l% q* T, R; a; [    Combining the results of smaller instances to solve the larger problem.) w0 L1 w8 T" }9 P$ W2 r8 n6 h
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Components of a Recursive Function
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6 r$ w' ^- A- {( j. R! i    Base Case:- m8 y0 S" T# z) u' M( S

- x; W, \% M# o. f( }0 C6 d# N        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.
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5 L* {' P7 R* |/ v8 J. a% b        It acts as the stopping condition to prevent infinite recursion.4 C. u% V$ R& i( W

" _/ [% |. r2 X7 e4 w! i        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.8 N* C8 `1 D1 g' s0 r
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        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).
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+ v, X' V# y4 V/ o# d/ DExample: Factorial Calculation; C  ?, @9 `' \$ F8 E
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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:/ a0 S! K1 g# x# Y

) Y$ ~5 @# t' _" z  x- O2 \# A+ d  M    Base case: 0! = 1
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    Recursive case: n! = n * (n-1)!
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Here’s how it looks in code (Python):
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+ Z! e7 y/ r* w# N: _# t, Udef factorial(n):8 S& F; F) G$ P" {+ Z0 w/ T
    # Base case( V) {+ u8 K7 w
    if n == 0:
4 ]. V3 Z9 ^8 }+ C9 R        return 1$ h3 y0 n; @& V8 h2 n  t. a
    # Recursive case
6 W, n0 f: {8 c6 h# Q    else:
' T5 `) ?9 ^" c9 f/ u. R% T$ _        return n * factorial(n - 1)' ^0 a5 I- c& k+ u# h
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# Example usage: P$ e& i+ \) B; R5 s# u* q: V
print(factorial(5))  # Output: 120* r  U. ^& D2 _1 D  V4 s

5 W; j+ T/ h' ~& s9 u$ e- oHow Recursion Works9 I6 ~5 M. X% O* a$ q

6 e; ]4 N, t4 j/ f9 M    The function keeps calling itself with smaller inputs until it reaches the base case." ~$ L; y% n# r# A" ?
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    Once the base case is reached, the function starts returning values back up the call stack.
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/ o( e# a: a0 v' @0 N4 {6 ]0 {    These returned values are combined to produce the final result.7 _- o6 x* X/ e6 T0 y
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For factorial(5):. Q4 j- I+ z2 ^+ ~4 ~+ ~
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+ @- L8 ], r/ d: Ifactorial(5) = 5 * factorial(4)3 P) B+ z5 p+ C* s+ n' M( S
factorial(4) = 4 * factorial(3)
  `4 u1 G! G4 ~0 {/ U2 yfactorial(3) = 3 * factorial(2)
/ E% l! u% [  M. _/ z6 ~: ~6 Efactorial(2) = 2 * factorial(1)9 M1 U& s) R8 q* ~4 Z( L- _9 q
factorial(1) = 1 * factorial(0)
$ A8 i/ n# T2 g$ x: |( K9 dfactorial(0) = 1  # Base case8 ^. ?) }* w6 q# n7 F
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Then, the results are combined:& S) H2 p: U1 |, @: w9 p
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factorial(1) = 1 * 1 = 1, W1 p$ K+ r- W+ g, ^! B  y( G/ G
factorial(2) = 2 * 1 = 2' C3 ]  P4 \) `1 @0 P+ f+ \
factorial(3) = 3 * 2 = 6
  ], @" H. ]3 L4 y" Rfactorial(4) = 4 * 6 = 24+ V" R8 t; Y- q" u$ J
factorial(5) = 5 * 24 = 120) R' x; ?" B8 }! y7 `5 j- P2 {. \

2 L6 `+ ^/ r# ^( d: iAdvantages 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).: a" ]+ ~9 c4 h; j( g! c5 Z0 ?* v

# `& f& x( H: \. C4 |    Readability: Recursive code can be more readable and concise compared to iterative solutions.1 r) o, M! ^5 K

; o8 D8 |4 D. @. jDisadvantages 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." [  R' v- O4 _# _
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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  E0 C3 k2 t" P# l3 T  i" V
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    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).: `/ i6 ?- K9 F  [, C; ~& q
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    Problems with a clear base case and recursive case.
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Example: Fibonacci Sequence6 `/ g  @  i) ~% |0 ]9 g1 p
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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  h# S% H  n3 W2 m1 b" `% [
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    Recursive case: fib(n) = fib(n-1) + fib(n-2)/ C7 |6 h" h2 q  ~7 B  p% z- J

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def fibonacci(n):
4 ^7 V/ Q* s$ o7 N    # Base cases
9 U8 u; h( a' ?4 n    if n == 0:4 E0 |3 D) f" v0 U- o* `
        return 0
' ^- {3 H. p: b2 D$ _, I2 _8 Y    elif n == 1:, }* E  F3 w2 t! s" ^
        return 1/ o4 ^  ^9 [; Z4 \# |
    # Recursive case, N, r7 Q% J% v* ?
    else:4 S' |  c9 r# ^0 ~9 X
        return fibonacci(n - 1) + fibonacci(n - 2)* s2 v4 Z2 F3 l2 o6 V  p. Z: H5 f

5 U6 t# }! p! s4 W" _: u! @, r# Example usage6 \0 Y8 i# Q: u$ @* T- r
print(fibonacci(6))  # Output: 8; x1 q$ v$ c2 V4 e6 G; d
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Tail Recursion/ ^. z% H9 {, h" ?/ j) j
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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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( ?8 x1 I; r* u3 N" C3 \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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