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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
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( r( n  I$ d$ S$ H0 J( h0 }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。$ G: }+ t# e8 G

) {$ V" O8 i. k8 z 关键要素6 T3 }& {8 K5 u3 A8 v; H
1. **基线条件(Base Case)**0 [: v0 U, Y& @" \4 Y0 i* ]6 d
   - 递归终止的条件,防止无限循环; g) D" d/ a# V' ?$ S: D' |( n5 N7 {
   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1
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/ |' J5 R9 x$ h" z2. **递归条件(Recursive Case)**; J, S; `4 ]+ a0 y* p  Z
   - 将原问题分解为更小的子问题. r8 n% z5 s4 J( \
   - 例如:n! = n × (n-1)!, l( f' Q9 c; ], ~( u, t9 K" J
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经典示例:计算阶乘; m  a0 w/ F8 C" Z) s
python  y. T9 O9 F: i9 {4 J( ~
def factorial(n):) Z5 |& m$ n. U% B  o: b
    if n == 0:        # 基线条件8 i' C( y' L4 \( Q
        return 1
  \. z/ y4 y- `% L7 p    else:             # 递归条件
0 ]0 Q% D; e1 S9 q8 m& F2 e0 [        return n * factorial(n-1)
) h, s( Z- j" x2 k执行过程(以计算 3! 为例):- L6 A) S' [: r
factorial(3)
5 H+ s/ U* v$ N2 B3 h4 ~3 O! l3 * factorial(2)
  H% W$ ?0 J" G3 N( e3 * (2 * factorial(1))4 z5 x8 A" A8 g6 C$ Y" o
3 * (2 * (1 * factorial(0)))
5 n: g3 h! H3 V) e( L3 * (2 * (1 * 1)) = 6
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递归思维要点
6 Z3 N" t3 [' D6 l0 z, z- S( k1. **信任递归**:假设子问题已经解决,专注当前层逻辑
% |: Z: s& L5 D/ ^2. **栈结构**:每次调用都会创建新的栈帧(内存空间)9 A, p0 w, n: M8 W
3. **递推过程**:不断向下分解问题(递)* f8 d; t; N! E# @; Y# l1 @
4. **回溯过程**:组合子问题结果返回(归), E/ c, J1 ?8 z! E

, h" z2 O4 ]4 j 注意事项. `3 Y! g" H- W
必须要有终止条件7 {2 j6 s* c* l  B) c
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
( c% V2 j# F! T0 D某些问题用递归更直观(如树遍历),但效率可能不如迭代
1 e/ I: M- G( _. @7 G尾递归优化可以提升效率(但Python不支持)
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# G1 S6 P& ^1 g0 M, Y& P 递归 vs 迭代
2 S* Z$ s  o+ i* E4 o|          | 递归                          | 迭代               |
/ u- T; H' U# Z5 [: x" X" E/ M4 {|----------|-----------------------------|------------------|
7 g* |1 S; _" U( J7 p( }| 实现方式    | 函数自调用                        | 循环结构            |% L; X" n" e% p& K  h( Y( S, h9 u
| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
$ e; L) r: p) T2 T/ h  R5 }! A| 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         |
$ c8 K; \! [, x- [) u' P| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |5 v8 E8 L& B- f5 H4 v  L' ?. h
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经典递归应用场景1 s2 v- z0 p  s7 `) z9 p; N: I* g
1. 文件系统遍历(目录树结构)* n3 d7 {8 `! v( ^
2. 快速排序/归并排序算法
: E. D! q" S. P& {* Q1 y2 y/ m3. 汉诺塔问题: f) T6 w. d6 h. `" W$ r
4. 二叉树遍历(前序/中序/后序)
2 u+ r, m, D; J" @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,
4 m, ~8 p6 T3 {% j& ?6 e) N我推理机的核心算法应该是二叉树遍历的变种。8 B5 K) }6 `$ M7 O1 x/ Q
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:
# Y: h. r& f3 v/ |% _8 x8 M# Y* fKey Idea of Recursion
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) J  L# N$ }7 l* {- Z8 UA recursive function solves a problem by:: J4 `: j4 \7 B& ^! A
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    Breaking the problem into smaller instances of the same problem.
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7 ?" d  A2 l: z6 }4 `8 F3 u. e    Solving the smallest instance directly (base case).3 B$ i# v9 f* c8 n, t

% k7 L2 ?# b* I6 z) x" [    Combining the results of smaller instances to solve the larger problem.
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Components of a Recursive Function, x5 |% R$ `- l0 s, b# _
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    Base Case:' q1 @7 t1 f4 Q7 h; h0 T

+ U0 n; B" k2 T        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.
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( ^/ [; o4 C2 q( P3 u5 k        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.- A7 q6 J  O9 X$ ~- E7 i& B

( a* i7 J# e. V1 P6 Z3 [# z; Y    Recursive Case:, [, X7 z& ^1 M/ k' ]8 K
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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% ?# p- ~7 d+ K
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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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% ]9 R: v; g, q; N! \3 a3 I    Base case: 0! = 1
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    Recursive case: n! = n * (n-1)!
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  n; _6 C2 k7 I7 T  i$ }Here’s how it looks in code (Python):
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0 l3 o  n: U; jdef factorial(n):
, `/ R/ ]; A5 Z) _. Y( o9 D    # Base case
# {# n4 Y! g. v    if n == 0:
; d- z" [  ~+ o  E        return 1' @0 R; K' `- D3 u7 V2 |/ t
    # Recursive case$ G, R; L: ~! i
    else:: a. F, F1 `: ?* N2 O
        return n * factorial(n - 1)5 _  S( v4 T) V& B8 L

) {3 {# O- y" `8 O& |# Example usage
& `% r5 W: V: L: qprint(factorial(5))  # Output: 120
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- _' N; q$ n& E1 S4 V9 {How Recursion Works
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    The function keeps calling itself with smaller inputs until it reaches the base case., B3 V: d! W( b* Q# P

) q( T8 ]/ a4 \# t    Once the base case is reached, the function starts returning values back up the call stack.
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" `- k) C6 F' X5 C/ Z8 P    These returned values are combined to produce the final result.
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+ F. G' o. j, c& h8 {  Z  K. gFor factorial(5):
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factorial(5) = 5 * factorial(4). h9 {4 P% \, k
factorial(4) = 4 * factorial(3)
. y4 i8 n% p& N3 J) ]1 L( mfactorial(3) = 3 * factorial(2)
& G6 W* X9 S9 a- z% afactorial(2) = 2 * factorial(1)
/ n& l6 o2 }/ s2 z. ]* ^2 Y; Nfactorial(1) = 1 * factorial(0)# V. C+ D0 x8 y* X
factorial(0) = 1  # Base case8 V6 P2 H# ?6 x
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Then, the results are combined:1 o8 Q2 R$ G: E2 v6 o3 E+ G

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factorial(1) = 1 * 1 = 1
% R+ O3 Y) j8 K! P/ @5 P3 Ffactorial(2) = 2 * 1 = 2- V9 ?) T% m4 @
factorial(3) = 3 * 2 = 6
( R$ ?2 {. l5 I2 ^factorial(4) = 4 * 6 = 24: q0 t9 o7 ?; R- J4 a/ }5 i! |7 f
factorial(5) = 5 * 24 = 120- L% W+ v' e8 V% y
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Advantages of Recursion7 B# B4 H" D1 e* L

/ r! Y, G9 r/ J- ^    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.
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* P6 X- X7 `) g" L4 ODisadvantages of Recursion2 D" ]) W' J) ?) u* |2 R8 ^# E

  m8 ], M3 K' i- S( \  H8 }    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.4 U, k% h( Q, D( g* V* a6 G
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    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).+ I% ?3 E4 ]- N) i7 w2 N

6 h! `3 i' g# K# u5 Q: A2 Q  D/ h1 y- ~When to Use Recursion
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* ]3 M& \, q( X4 C6 t    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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$ U% J5 }8 t/ _Example: Fibonacci Sequence
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  ^( F1 l' G& k  _The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:2 ]( u) Z/ Z% }# M9 @- n

0 y3 B3 Y4 U& ]& _7 P* K* P9 w) f0 s    Base case: fib(0) = 0, fib(1) = 1) \5 X. l. c+ t9 z+ v; \

% {1 j) i3 N" U7 ^    Recursive case: fib(n) = fib(n-1) + fib(n-2)4 R1 R, E; \+ [$ n; ^% H  }$ P/ G

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def fibonacci(n):
4 ?$ {7 _# e  Y- ^2 e# F; p3 b  `    # Base cases
* W1 j0 R  h) D6 g) H9 H! ?    if n == 0:1 Y0 b2 y) l$ W
        return 0
; D+ g; B9 u& N" I3 \% T1 u) H% f    elif n == 1:
$ k& L/ J) r4 U/ S- h& K        return 1& P- c7 e" a" u8 b
    # Recursive case, d' R  e" a# M, ?
    else:
( L2 N& {% R7 J! |4 x( y: [        return fibonacci(n - 1) + fibonacci(n - 2)
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  O1 |' `& |# N* V# Example usage
4 i5 E; [- R7 |print(fibonacci(6))  # Output: 84 H3 ^) J! j; T% ~- Q' `

1 T: N9 j6 i: eTail Recursion
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" ]& W/ M; F. w& T- [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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