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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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; }" E: Y! E1 v4 L, u解释的不错
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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。- Y0 P! ]* E8 L
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关键要素
. I% C# s0 }* A9 D& j6 W% Y% l1. **基线条件(Base Case)**2 c+ H1 T1 h: L( U- h- R
   - 递归终止的条件,防止无限循环- ]3 z3 h, Y7 i1 U0 @
   - 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)**
/ l  {/ e" p+ f- W* f$ _   - 将原问题分解为更小的子问题
- E7 s' [' E0 @( A4 t$ z$ K   - 例如:n! = n × (n-1)!
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经典示例:计算阶乘
# E, _2 `& _  \( C) Tpython
4 b9 |/ K: k# D4 s3 D% @: {! Edef factorial(n):% N4 M( M, e- d3 w) h7 J
    if n == 0:        # 基线条件3 H* U2 ^% [- @4 u3 y6 \1 s
        return 1! L7 B& p) A8 w9 Y6 h/ W8 f
    else:             # 递归条件
  b4 [. n9 b* ^# N! {3 K        return n * factorial(n-1)
, G/ E2 }2 Y. e: x9 v. |. G- M  [* L执行过程(以计算 3! 为例):/ M) M5 i9 q/ k6 l; E8 \$ s
factorial(3)
" t6 U# k# {$ y! G3 * factorial(2)
# s/ V+ Y3 V* M6 C0 N. X2 Y* _8 w3 * (2 * factorial(1))
9 |! W" L7 E+ [! H1 P3 * (2 * (1 * factorial(0)))  s% r) e! }4 \4 g- S
3 * (2 * (1 * 1)) = 6
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% E$ H) `; X9 b 递归思维要点
$ @) ]* N5 C6 P# G% }5 L! `; d1. **信任递归**:假设子问题已经解决,专注当前层逻辑7 B7 _9 s, M" K, q1 y
2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
7 E3 P" s& a/ i( Z2 @7 H6 z4 N3. **递推过程**:不断向下分解问题(递)
5 y. F$ @7 y- \9 Q1 j5 u4. **回溯过程**:组合子问题结果返回(归)2 r/ |; t: }8 F; @9 Y$ Z

: u/ m9 y4 i  i9 T6 _: \ 注意事项0 d9 B( G' b; ]9 q
必须要有终止条件# D- B; n5 Y/ i& T0 [
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)) f; ^3 j# M' Y+ A1 B0 q
某些问题用递归更直观(如树遍历),但效率可能不如迭代1 n) M: P6 s2 l2 d8 }) {$ @
尾递归优化可以提升效率(但Python不支持)4 n7 u2 K$ ~: M

' a$ Z; W: p" |+ v3 \) M* B% i0 F 递归 vs 迭代" C. K4 L. D5 F1 U5 U
|          | 递归                          | 迭代               |0 ^" ]! v5 M9 S/ h  j1 j
|----------|-----------------------------|------------------|
2 S) Y  M+ ]# g  ?( r0 u2 q* |& d| 实现方式    | 函数自调用                        | 循环结构            |
: }" e5 M$ Z# L( G1 v& @7 C( l0 M| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
3 @' ]/ I3 y8 q7 {( ]| 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         |+ [! N: s2 P; {# L: ^
| 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
5 M2 L, M- ~9 P7 G4 g1 I1. 文件系统遍历(目录树结构)" \% e4 h; j1 f7 H
2. 快速排序/归并排序算法( J3 h# x% j! t. d; r- I4 f; @- U
3. 汉诺塔问题
6 N8 d9 l# W! X4. 二叉树遍历(前序/中序/后序)) @( d* M  a, c2 R5 k: g4 X
5. 生成所有可能的组合(回溯算法)3 h1 W' Q0 X  _! G1 P8 N8 f
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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 G& @" X) R+ r3 e# Y我推理机的核心算法应该是二叉树遍历的变种。! d: q' i5 d% b! 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:
1 I& C! [3 F+ B+ B0 KKey Idea of Recursion
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3 ^, E- v" @( t+ r7 F. T8 c; gA recursive function solves a problem by:
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) g) Z" _8 D1 T/ y9 p  o6 \8 t    Breaking the problem into smaller instances of the same problem.
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    Solving the smallest instance directly (base case).8 ^1 B4 G( B; w, s$ z; i

% ^) P8 x) y6 u* i1 u) }    Combining the results of smaller instances to solve the larger problem.0 ?# ]3 o( W. s$ [: m" S+ Y9 ?! M

7 g" z* i& K( p  |2 {2 G7 b8 R3 OComponents of a Recursive Function, D# L# _+ `( B$ {* R) h

* p& M/ x4 f5 _1 X6 o    Base Case:
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" `" @  ^/ w0 d5 G, k  b& q& `$ J        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.4 m( N( p5 v' R3 t; `
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        It acts as the stopping condition to prevent infinite recursion.
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# ]- k$ {/ g) t# H, j9 Z  p0 Z        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.
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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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5 \  i& `. w* H" Y' y, P5 [+ {5 Y; r8 kThe 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:+ K8 r' o1 R# u8 L! \

3 H9 y# e% T4 P0 V3 y) W  _$ K    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):/ N* X1 J3 W$ n. J3 g
python' b6 r* v- M% e0 j7 ]7 J

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def factorial(n):
$ {. \& f$ m3 \* Z! _2 j    # Base case
# J! I. }9 W) ]- m    if n == 0:
" u4 C9 j5 u+ o        return 1
5 w+ k) }7 `7 H( m# b+ \    # Recursive case
; h( I- G. j9 [" T/ X3 g    else:
; y- L) D: N3 x8 I        return n * factorial(n - 1); p, g* B3 n! g( q7 |* H

/ P; h- f& z3 i" ~6 [# Example usage
- J( ?, g+ g* p+ C8 {print(factorial(5))  # Output: 120
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How Recursion Works, @& f. T# O, I7 v, r3 f
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    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.0 y. n# k3 b7 |& A" b! t
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    These returned values are combined to produce the final result.+ _1 s0 @% T+ i1 Z6 ^9 @

1 u' q2 l% j( lFor factorial(5):
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: e$ d  x- P* V" Mfactorial(5) = 5 * factorial(4)
+ U/ X; C0 H$ v9 zfactorial(4) = 4 * factorial(3)
% {3 a4 T) @% y$ I- Rfactorial(3) = 3 * factorial(2)# Q  p( G+ p. e. _
factorial(2) = 2 * factorial(1)- o8 K/ Y$ `" `# ~% ^
factorial(1) = 1 * factorial(0)
% V1 g% |. B  _' rfactorial(0) = 1  # Base case7 l$ P7 o# z3 j% c, E
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Then, the results are combined:
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factorial(1) = 1 * 1 = 1, E, t3 R% e, }$ _" Z. b
factorial(2) = 2 * 1 = 2- J7 u) N7 b8 W  [  x, n
factorial(3) = 3 * 2 = 6: i8 r( x- u- ?+ T) a  M
factorial(4) = 4 * 6 = 24. C' c" f- n4 S
factorial(5) = 5 * 24 = 120
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& I# t) g1 {- B# T$ n3 R6 IAdvantages of Recursion
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+ }. w9 D$ K1 Y/ j7 N, p    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).: O/ N8 t; s3 |" Q
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    Readability: Recursive code can be more readable and concise compared to iterative solutions.* R- r7 C" W5 g4 d: R

3 ^, [/ U$ w3 Q5 wDisadvantages of Recursion' v3 ~- b% l9 S' A& b) v& B, k" G; Y

2 L  `: o- k% F. I+ i: ^    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).( S" u3 _$ e+ A# p7 N! X+ @, |1 ^. V
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When to Use Recursion
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  c7 g( M) p; O/ x* G    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., I4 i; G  f9 d" y, Q, ^4 e# }
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Example: Fibonacci Sequence' \7 i- C2 U: ^3 i
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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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, h6 J0 y( i  ~4 |2 F    Base case: fib(0) = 0, fib(1) = 1) ^$ d( `  V* C* H; L' F
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    Recursive case: fib(n) = fib(n-1) + fib(n-2)
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def fibonacci(n):
7 B2 |9 ]' c1 h. V    # Base cases1 o- f5 c4 ]. f& g) n, x
    if n == 0:3 N! }3 ]1 E4 u% d
        return 03 O  c8 l( H7 N( V$ i
    elif n == 1:
8 o! L9 k: h) H, I        return 15 M# V" [# d% K$ S  b9 T
    # Recursive case
7 w$ x1 n+ G/ N  f6 L) P    else:, p3 C  X, b. H" e% o
        return fibonacci(n - 1) + fibonacci(n - 2)
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+ s$ j! d! o5 v; z4 K) C$ A/ B# Example usage  e3 E! L1 P6 m# ^9 W7 W2 x
print(fibonacci(6))  # Output: 8- x- g- n3 R& O  L' X4 p, l

2 I- W% U- _9 e; mTail Recursion
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. E, q7 g' r5 F* \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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( ^; P  b' O. t7 y8 L- @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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