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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 ' E3 @# _$ d) S( P! V8 ]
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解释的不错- @* B( X. W& k' J4 z
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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
+ ?; H% M8 C  ^$ |( x' o2 }1. **基线条件(Base Case)**0 @' r$ r# b4 R1 f  J! ^) M+ \
   - 递归终止的条件,防止无限循环7 P( e* E- O# c# P0 R
   - 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)**
+ c( D+ q* X" q2 g7 t$ ^' ?   - 将原问题分解为更小的子问题
  Y% }  ?& {, w0 j* v# V' P( n   - 例如:n! = n × (n-1)!
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8 ]5 p' x) `3 y, I 经典示例:计算阶乘
  }8 x) O9 h& C; \( q; C" Y9 P4 Npython
9 L& ~  y" ?1 z8 }def factorial(n):
0 ^; Q7 |7 p! J) k: q; \4 A    if n == 0:        # 基线条件( g  ]" X! o# u: @3 U4 ^, m- G0 W
        return 1
- a, a2 ~- n3 b4 ]( G1 H    else:             # 递归条件
, ]) q1 v1 A3 o, l$ m        return n * factorial(n-1)5 O/ v" Q' G" u9 ?
执行过程(以计算 3! 为例):
3 z" P' S* \9 a. Ofactorial(3)
! T1 A( v: U3 Z  E0 w3 * factorial(2)
* v0 j" ~; P9 {0 e3 * (2 * factorial(1))  p% V* O9 u1 n  i$ ~
3 * (2 * (1 * factorial(0)))+ {, ]  ~- l% w* s" r" {+ {4 O2 x, s
3 * (2 * (1 * 1)) = 6
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* d5 B- s" ]7 ^, Y 递归思维要点9 M4 p4 h/ |: z( n1 y
1. **信任递归**:假设子问题已经解决,专注当前层逻辑
) ~0 u0 u& e; X' O3 D% ?2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
% S/ g; }. B2 w2 n" q& ^: x5 P0 a3. **递推过程**:不断向下分解问题(递)" Q4 y0 s4 }, ?
4. **回溯过程**:组合子问题结果返回(归); A! s) b' [3 q! A
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注意事项" c$ G) D( e1 v3 ]5 u2 V% d
必须要有终止条件, H, f/ ]$ p$ l( _( I% F
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)8 I' L0 P& M9 R0 K8 Q) R. x9 L
某些问题用递归更直观(如树遍历),但效率可能不如迭代, a" Y& R: K* e7 e; j5 |
尾递归优化可以提升效率(但Python不支持)3 [* V7 s" R$ C) u7 C

6 S! ^9 E! ~0 J5 H' U3 B 递归 vs 迭代/ p8 J, r) w% I1 L* I" m
|          | 递归                          | 迭代               |! \0 a5 x4 d  Z* w% m, c9 g! r
|----------|-----------------------------|------------------|
2 X% s: R2 g6 z| 实现方式    | 函数自调用                        | 循环结构            |
3 V4 m# I' @% Z" c5 k) P| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |9 V$ Z& E4 A4 H6 E
| 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         |# f/ f5 F! V& d4 V( P: F4 m0 B
| 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
/ l+ I) |! V) q1. 文件系统遍历(目录树结构); N+ f) @) W6 v% A
2. 快速排序/归并排序算法
- q4 k# a8 |- c" S0 Y: y3. 汉诺塔问题; g9 m' }- d$ d* b& Z5 l$ ]! I
4. 二叉树遍历(前序/中序/后序)5 ~! a+ u3 g2 ]
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,2 J: R6 y! x- {
我推理机的核心算法应该是二叉树遍历的变种。; A# s4 I3 A4 {5 b2 \
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:4 R6 s( k1 i) ^! z
Key Idea of Recursion' d. Y+ N8 I, a: o: \+ @0 Q8 @

/ S; h3 h& _9 H9 o6 L1 `A recursive function solves a problem by:
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7 }9 t0 ^6 ~) N5 }    Breaking the problem into smaller instances of the same problem.
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5 Y" q: V- ~$ ~) J    Solving the smallest instance directly (base case).7 E& n# X- [+ K1 q# w) [$ V# U
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    Combining the results of smaller instances to solve the larger problem.
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& c1 w0 g% R. S5 nComponents of a Recursive Function
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( l: x. r7 d0 V    Base Case:
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, X! y% t. p1 V        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.
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- Q5 {% X# a' B: X6 a6 d        It acts as the stopping condition to prevent infinite recursion., V, n' c0 M9 W$ ^  P  |
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        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.
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+ C2 b  |+ R0 Y$ a* V5 E+ t5 R    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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4 K* B1 }0 _( x- s7 \& ~        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).$ f- S  [! l0 u

% d4 W! U9 h  Y, H  f1 m/ LExample: Factorial Calculation
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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:: J8 y5 l* _6 T! P$ t* G' R5 {' E4 u
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    Base case: 0! = 1
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3 a# ~8 z) c+ j    Recursive case: n! = n * (n-1)!1 Y# g& X/ R/ g/ d3 a

* q( A) I0 w, m' cHere’s how it looks in code (Python):
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  C) C( S2 D/ jdef factorial(n):& t' o2 ]+ q6 P2 K; r
    # Base case# @9 M; l* e+ R- E4 N- a
    if n == 0:
" p  _$ a3 ?( z$ i' `        return 1
4 U& {5 U: ~$ c; q8 V  ~    # Recursive case# Z& ]0 e/ w4 i, \- ]
    else:  _; Q9 f) d: K# M9 h  I
        return n * factorial(n - 1)
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# Example usage
% |' \  R9 h/ dprint(factorial(5))  # Output: 120! a+ b, e6 _+ H+ N. ?0 c; Z

! z5 h% y2 t5 [: m9 v  p& C/ i% P2 vHow Recursion Works
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    The function keeps calling itself with smaller inputs until it reaches the base case.! i( W+ B% _% {3 n) O( Z

3 \# y6 p) }# j6 C" e2 c9 X% a: K6 L    Once the base case is reached, the function starts returning values back up the call stack.
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    These returned values are combined to produce the final result.
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7 e  N: z2 L) ~0 K+ D0 U, \For factorial(5):
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1 ~2 p; F2 y( r& dfactorial(5) = 5 * factorial(4)
/ o/ U, q: J: {3 }: F8 `factorial(4) = 4 * factorial(3)! b* t- W- v+ o6 d( Y2 K" \3 ^' @
factorial(3) = 3 * factorial(2)
6 i5 H6 @: C) u' ^% Z6 Z- e' ofactorial(2) = 2 * factorial(1)) {  w  T% k7 |& E; c
factorial(1) = 1 * factorial(0)
: h: Y1 I% M  ~! a( H  G& Wfactorial(0) = 1  # Base case- H8 w; J" U7 i* R3 n- B5 T) t

$ T2 f3 ]0 Q, D, c+ lThen, the results are combined:
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factorial(1) = 1 * 1 = 1. V$ N8 q$ x* x% _- O6 C  r% }( I% \
factorial(2) = 2 * 1 = 2
; I2 n5 `7 `. z/ qfactorial(3) = 3 * 2 = 6/ X, L# t9 s2 a9 A1 p
factorial(4) = 4 * 6 = 24, l# l4 @. Y. `( H/ Q
factorial(5) = 5 * 24 = 120% D5 K; n1 D8 ~+ L( J. t# O
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Advantages of Recursion
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5 m" k0 l. Z: O) X  L/ i4 o1 a  R. |" q4 h    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).& q( }7 {, w7 {9 j% q% W. L; i4 |6 \. \
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    Readability: Recursive code can be more readable and concise compared to iterative solutions.7 N$ y, ~& r/ C" g( [% S
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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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) P5 ^" {( ~4 E* H    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
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2 O0 `/ G- F; C, B% `When to Use Recursion1 n3 {/ o4 M$ R2 s  Y4 L' ^

* _# }8 Y+ i3 z- f! h    Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort).
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3 q7 s$ |$ t: M- p, y* O    Problems with a clear base case and recursive case.( I, ~, x# ~1 q1 B
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Example: Fibonacci Sequence3 \- }6 q* j. Y  w' g$ `
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The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:; m/ }2 ~% f  t- i

) `8 [' h" T/ ^. n, T    Base case: fib(0) = 0, fib(1) = 1, M- C! x+ w' [, m0 `

  `- b# @- N* d+ h3 a+ p1 b    Recursive case: fib(n) = fib(n-1) + fib(n-2)
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def fibonacci(n):0 b0 w" L2 [4 R' |% w1 ]0 X
    # Base cases$ f& \8 d; e  k! u" J+ i  M: z$ E
    if n == 0:
: m7 ]  r5 @+ ]$ g2 z; u! j        return 0( O6 J* E; H  z) v
    elif n == 1:
! r; g3 I! H; A! B' X- `- s* @        return 1
/ P  h' a5 S( D    # Recursive case
& {% v9 H% D4 j5 @* s) m    else:2 S  m% h. _/ ^
        return fibonacci(n - 1) + fibonacci(n - 2)4 D' b: }  o0 E7 T
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# Example usage
9 N; T, s# w% x/ u7 Hprint(fibonacci(6))  # Output: 8
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Tail Recursion
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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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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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