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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 ; M, v2 I$ ~5 S6 G( F. ]6 r( z

( F7 C. m2 k: h$ W解释的不错' Q& s3 {% F; U# u

' E3 w2 h% v/ f( ]3 d& [3 d# t1 l! w3 }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。( ~5 h3 V$ `& g2 R9 L
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关键要素7 k8 G$ x4 }) A+ ~! {' V" D6 P
1. **基线条件(Base Case)**; S7 @. y! N$ ^2 d
   - 递归终止的条件,防止无限循环
" h  k4 N. L1 D8 a" \- l5 v   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1) d3 j9 y: B: S- Q1 C
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2. **递归条件(Recursive Case)**% H7 p( a; V: E; G$ }  i) D: E
   - 将原问题分解为更小的子问题0 f1 K2 ]& E# }1 |* w* {
   - 例如:n! = n × (n-1)!
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1 \# `/ i" N4 n5 A, I. O7 e6 u 经典示例:计算阶乘
$ X* U3 X" c! a: v) X3 L/ V5 q8 i, F  bpython
* P; Q& }8 ~( O$ `3 b2 udef factorial(n):* |! ^5 ^# x9 E
    if n == 0:        # 基线条件  M5 ]  E; x1 Z9 @
        return 1
  @8 F, _3 l7 N- E    else:             # 递归条件
( G4 w, l- g6 H5 A2 y" O; y        return n * factorial(n-1)6 Y1 G0 O# f  y( K/ B
执行过程(以计算 3! 为例):# m- P7 L$ H# f2 f6 `% V
factorial(3)
( U+ d  L, q" x! D3 * factorial(2)/ u+ \/ _! ^# A/ U
3 * (2 * factorial(1))
; }3 f! Y8 C! V' v" J# s3 * (2 * (1 * factorial(0)))
" N8 P* m9 }# V- O3 * (2 * (1 * 1)) = 6
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递归思维要点
0 f# j" g$ z( z1. **信任递归**:假设子问题已经解决,专注当前层逻辑
6 U; N8 V9 g, X# \2. **栈结构**:每次调用都会创建新的栈帧(内存空间): `0 G9 M' n- ?' o! }3 w
3. **递推过程**:不断向下分解问题(递)
) X0 S2 v+ k3 u; G4 ?* m( x% ?9 p/ Z4. **回溯过程**:组合子问题结果返回(归)! Y3 A2 y  o0 k; m# p
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注意事项: e( G/ k3 j8 S0 I, x
必须要有终止条件3 r' d+ q9 [) g- d, i
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)$ L; _; i' M, X- k6 V
某些问题用递归更直观(如树遍历),但效率可能不如迭代# f) z9 K  _9 k0 Z1 ?* c
尾递归优化可以提升效率(但Python不支持)
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8 }- B% [. F5 p* m9 J% C 递归 vs 迭代
0 O4 \& o, I7 ~8 U# J2 u! r) ?|          | 递归                          | 迭代               |) r5 j% I6 r1 A% w+ d
|----------|-----------------------------|------------------|
5 B* U$ B4 G4 }| 实现方式    | 函数自调用                        | 循环结构            |/ [% \/ @1 S  x( K; ]6 g: z" S
| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
7 v3 j( U$ \+ w3 ~4 Y| 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         |- H5 G2 e- Q; f# ~* @4 N3 |8 x" \6 Z
| 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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2 r5 N$ ~; E. G. d5 X& A- K4 O 经典递归应用场景! g, o' J' U5 k" c% p; ^/ |4 M
1. 文件系统遍历(目录树结构)1 q8 C# e1 C5 W* N# x- y6 ^
2. 快速排序/归并排序算法' R% W8 R5 M* @5 {, G  q
3. 汉诺塔问题- B5 d) Z6 `  w! r6 O3 y5 y' P, M
4. 二叉树遍历(前序/中序/后序)8 Q+ |# [; i# i8 i7 r
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 ]5 ?) N( e+ N我推理机的核心算法应该是二叉树遍历的变种。
, o$ ^1 m! S  g% W5 |5 g% c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+ f' p2 g$ y5 WKey Idea of Recursion' T+ e- Q7 A0 M( l2 z; q
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A recursive function solves a problem by:, ~$ \7 T- h/ Z0 ~$ Z4 B+ k8 ]

9 ]' p% F! Z3 T    Breaking the problem into smaller instances of the same problem.+ p% a" c  }/ d
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    Solving the smallest instance directly (base case).
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. z& d- j$ I' L3 o; F    Combining the results of smaller instances to solve the larger problem.
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* T2 l9 E* Z7 k9 ^; [) Y* R, VComponents of a Recursive Function
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& s" D! P) q' c8 w2 @9 c    Base Case:
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        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.
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        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.: A4 A( v5 Z: b6 s/ z; m/ M

  b& @+ n7 Q$ l    Recursive Case:4 O0 I9 y. _" I
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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).
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Example: Factorial Calculation
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2 w; v! O: C$ ?* g) b4 ]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:- R6 h, \" S# v" I% [

$ L( f% p1 I2 R6 R    Base case: 0! = 1+ l; o% y) T, M9 ^6 |
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    Recursive case: n! = n * (n-1)!
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Here’s how it looks in code (Python):3 {' ~; v* z8 `+ p5 V
python
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, L& U2 L) i8 g( {& wdef factorial(n):
7 e; i! g; O% o    # Base case6 }! Y* ]1 I" g$ V7 J, y) _& }
    if n == 0:
2 b  l( Q2 s1 H4 G) E" K6 M9 y        return 1$ y" \) I# P: }- G3 E. K1 P
    # Recursive case
% o2 q& G, {' @8 r    else:. N' _& h/ d0 o$ M/ B6 v5 n% t% |
        return n * factorial(n - 1)
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# Q+ [) G: o3 \# q. ]% B0 _( ~# Example usage: N% C' E3 ^+ I; x9 Q" k
print(factorial(5))  # Output: 1201 w9 X3 [9 L5 ^( S$ a7 q* h
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How Recursion Works
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8 }. H" H% h& L/ q    The function keeps calling itself with smaller inputs until it reaches the base case.
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7 N8 I/ t, n- J4 [% `# f! N    Once the base case is reached, the function starts returning values back up the call stack.
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7 @) q. p" U, M3 k. W$ w% `$ @    These returned values are combined to produce the final result.2 B, D8 x2 B: N5 z9 g) L
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For factorial(5):
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# C( }; ?* V4 g5 l  Y+ H. Q- F$ |) vfactorial(5) = 5 * factorial(4)
  a. Z! P9 `2 N* c& }2 t6 }factorial(4) = 4 * factorial(3)& L' Y1 b$ ^0 e" C* c& f2 p
factorial(3) = 3 * factorial(2)
6 x( B  _' N$ J) `/ tfactorial(2) = 2 * factorial(1)
) s7 D3 ^! a2 m) E# dfactorial(1) = 1 * factorial(0)9 {* r9 Q6 g! _( |( Y4 |6 l4 u( H) f
factorial(0) = 1  # Base case
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Then, the results are combined:+ P& p8 {& H3 i/ @" P. `

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, R" b5 v2 ~: f( j4 M. ifactorial(1) = 1 * 1 = 1
/ @. R/ \4 D' M( zfactorial(2) = 2 * 1 = 2
' u: h9 u  S# b4 n- Nfactorial(3) = 3 * 2 = 6* k6 d9 G% h1 m6 e, `
factorial(4) = 4 * 6 = 24
$ G2 U3 y# g3 \, d2 zfactorial(5) = 5 * 24 = 120
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Advantages of Recursion- g" s( }9 t6 B. ^. E$ }9 z- A

1 d& Z& b& V# M( v0 `, a    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)." h% M; y0 t$ t) Q$ N1 g( l+ Z
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    Readability: Recursive code can be more readable and concise compared to iterative solutions.
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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).4 ?$ b  g, ~( }: V* I" @5 t' ^# ^4 F; [
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When to Use Recursion2 y. c, t3 W; v$ g% r$ T
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    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).1 k4 ^. z/ j% Z+ I  _, d! e

' @/ U; N2 s2 a8 W7 l' @) H    Problems with a clear base case and recursive case.
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, W3 l" `0 m' _Example: Fibonacci Sequence
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6 f2 s" q2 K& w0 Q6 tThe Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
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2 _# q: q6 {$ e3 `    Base case: fib(0) = 0, fib(1) = 1
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    Recursive case: fib(n) = fib(n-1) + fib(n-2)
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0 [. D9 {, e* G0 Rpython
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5 _& P8 v3 h6 m# [def fibonacci(n):
- Z5 w1 V% Z( q4 k. O    # Base cases! ~5 I* Y" n& [/ u$ u. l) _% l& U4 e
    if n == 0:
9 X) I% O' G7 e* |0 b6 _) v        return 0/ ^1 B4 ~- O4 \
    elif n == 1:
# X( z4 ?" v7 \        return 1/ G1 J# ]9 i* H) A
    # Recursive case$ ^$ ~2 l! l) r7 ^* j5 b
    else:
+ h7 K2 E' y' t" k+ R        return fibonacci(n - 1) + fibonacci(n - 2)
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# Example usage
# |5 l. x$ A, f3 X# eprint(fibonacci(6))  # Output: 8
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; ^7 p, q) j1 a! c, `% J$ b7 r6 n3 kTail Recursion5 K, O9 T( X7 h$ c

" m& c$ p0 Q: \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).% q5 A! U; ]2 F* o
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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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