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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 % B# J8 E9 a+ F( a" R& A
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解释的不错/ S+ Q" M( M% X
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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  G8 S, j/ q0 ^  i2 T6 |
1. **基线条件(Base Case)**
4 p$ j7 I7 }; m8 Z: ]   - 递归终止的条件,防止无限循环% g0 ^" J& N$ B  c. `. F
   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1/ Q* z8 v9 ^+ w- C. [7 _. I* P

- A& P+ K- r# H4 A/ o2. **递归条件(Recursive Case)**" g8 q7 k2 }; @3 S- Z
   - 将原问题分解为更小的子问题; Z# p( {; l7 c2 T' M! Y
   - 例如:n! = n × (n-1)!! D5 }; f1 t& @6 `! x, f

% U. X+ r4 n7 a$ ^7 d! o/ E' f8 { 经典示例:计算阶乘% _7 w- d% C# N: ]
python! w$ U/ d) D. w: c5 t* E( E" P9 P
def factorial(n):4 c3 Q7 `' k8 ]& `  Y+ f
    if n == 0:        # 基线条件
( _6 K5 f7 k) o3 r8 }        return 1
5 [$ Q/ {, n; _' V: |    else:             # 递归条件
/ \$ D# L1 S) w8 ^! g3 l        return n * factorial(n-1)
. F1 E* o9 r( o2 k执行过程(以计算 3! 为例):4 W  r) L( }1 A9 x$ y
factorial(3)
# k9 U; f, o, M# g: {. c3 * factorial(2)* K  B$ Y, P$ M
3 * (2 * factorial(1))9 L4 Y# |/ {% V, F# M; S
3 * (2 * (1 * factorial(0)))
5 {# S9 {( u4 w% N7 N, x3 * (2 * (1 * 1)) = 6
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/ H  N, ^1 K1 @  W9 g4 R5 Q 递归思维要点
+ T. u/ @/ V. g3 d' @! L1 G9 j1. **信任递归**:假设子问题已经解决,专注当前层逻辑
% g, ?2 U" G! T; A2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
# Q1 y) W- T0 W4 C' e: N3 J  a1 l3. **递推过程**:不断向下分解问题(递). ^* b/ Q& e, o$ p7 I; ?
4. **回溯过程**:组合子问题结果返回(归)
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2 y+ W& q, B# U& m 注意事项
, q% G$ T! R/ w9 x% C. G必须要有终止条件1 I# x3 w7 Y% i/ r+ E* Z
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)# z) z& R* A0 q0 S% d" w" d4 V
某些问题用递归更直观(如树遍历),但效率可能不如迭代4 N6 N& {9 |: v1 ?
尾递归优化可以提升效率(但Python不支持)& O+ u6 d8 f# F% x" Z  L
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递归 vs 迭代
7 X6 ]. g, v% s) L|          | 递归                          | 迭代               |
$ B2 P7 R3 s& q& x0 q( c+ e|----------|-----------------------------|------------------|
2 E* g1 x$ L% A+ m. A8 p| 实现方式    | 函数自调用                        | 循环结构            |
3 K  `% g) [! v( L; F! r| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |) }9 ^0 t' y# g, f" H) k  b& o$ m+ Y" N
| 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         |5 J) t, r& i# d. A! `2 ?
| 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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7 P0 m2 r. d# b  P$ L. l1 [ 经典递归应用场景- N$ y4 [4 U/ v  }0 d0 a
1. 文件系统遍历(目录树结构)
8 g+ c2 {; s7 G2 z& L1 }  I2. 快速排序/归并排序算法
# }. p* f  V2 F0 h3 o5 X. m3. 汉诺塔问题
1 q) P4 ]( h; {  A) w4. 二叉树遍历(前序/中序/后序)
# u. X& ~7 m5 x, F5. 生成所有可能的组合(回溯算法)
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0 D" J9 C3 E6 e+ K6 ~% F- R  y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,
, k" Q; F) ]: M我推理机的核心算法应该是二叉树遍历的变种。3 D- U( `- j' u6 @) A* n
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:
) O1 A% w: @0 @5 q5 G/ B/ QKey Idea of Recursion
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5 {# V6 s9 u; G7 x: DA recursive function solves a problem by:  E; {7 V: w$ i/ A6 D

1 C7 J3 z/ E* L6 m; _0 u    Breaking the problem into smaller instances of the same problem.
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: w6 Z! c" F1 e: b' a$ Y# ]    Solving the smallest instance directly (base case).
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    Combining the results of smaller instances to solve the larger problem.
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6 n- T: |& r) zComponents of a Recursive Function* R0 @( y% |9 U7 P
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    Base Case:
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1 }# g2 V6 N6 }$ v& i) J, b        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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8 r0 u5 B3 X6 ?. z/ z, h        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.2 v8 f: w' F& _0 T  {2 y& B

0 X( h; h" s: Z3 m) }    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).6 Q% }6 L1 @" {+ S: L: {' b( {

( {3 m8 l0 z$ b9 EExample: Factorial Calculation! E9 @4 ^" A2 [) U

0 _( t/ i# M9 ]1 [! aThe 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:+ d. a  F- v: M; r
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    Base case: 0! = 1/ {. y/ z% u$ ~2 V6 |1 I

9 O, P& p% d$ ]. Z    Recursive case: n! = n * (n-1)!
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5 _) `5 c' ?$ ^; G+ NHere’s how it looks in code (Python):
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  a: R5 }- q; sdef factorial(n):% E- K: \4 |3 _7 }
    # Base case
& z  m* z' F- o* p$ d    if n == 0:( }$ @4 g; M" G0 J
        return 1
1 J* U$ W/ W- b    # Recursive case* U. R4 \4 d* P. Q! C
    else:
/ V, R$ f* Y% V( q  A; G        return n * factorial(n - 1)1 P8 a, a" U) D' }" `- ~

( E3 b5 `, p& a+ A" I# Example usage
" ^1 t  S; d! R2 a  ~0 pprint(factorial(5))  # Output: 120# J5 y! h. p- h# I5 ^( g; @

8 ?: k" z- @7 M! `+ OHow Recursion Works: X8 i$ R* i$ Z0 [& 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.
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4 }, V  L& q* j" y    These returned values are combined to produce the final result.1 ]" v% s' s  K7 g* h+ F  {
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For factorial(5):
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0 A' t! r5 [" ?/ m3 |( i. yfactorial(5) = 5 * factorial(4)
* p' y& Q" r2 O8 k/ Lfactorial(4) = 4 * factorial(3)
9 B: f9 j* K* h$ Y9 Y+ V9 bfactorial(3) = 3 * factorial(2)
/ m" s0 {, X$ Efactorial(2) = 2 * factorial(1)
% L9 t% _" H* b) m. ~4 f2 \2 b$ `factorial(1) = 1 * factorial(0)
; u- k, ~. x! _# i3 Ffactorial(0) = 1  # Base case
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Then, the results are combined:
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factorial(1) = 1 * 1 = 1$ O$ N. R2 ~. R
factorial(2) = 2 * 1 = 2
7 }. B+ s1 `/ L) K$ X) sfactorial(3) = 3 * 2 = 6
5 N- F2 r7 H6 V4 x: h) P" F" c3 tfactorial(4) = 4 * 6 = 24( R* ~7 j. d5 ~, w3 \( u' ~, s
factorial(5) = 5 * 24 = 120
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Advantages of Recursion
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& W/ i1 x7 I. D0 M, j+ F3 q" Q    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).' S7 n# }4 R3 ^) n

5 I$ T7 e; Z) r7 K. g    Readability: Recursive code can be more readable and concise compared to iterative solutions.0 g5 S7 G5 D8 M. @

9 |+ F7 E% f8 X2 fDisadvantages 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).& B$ o5 x! z8 c5 L  {. a( _: X
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When to Use Recursion2 F" H0 d- \) n1 Y
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    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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4 T# m8 J- L. M5 D, ^( tExample: Fibonacci Sequence$ v0 O) d. Y2 b" U8 _; x8 [: c. m

2 r) ~% e4 v; Q/ h0 ZThe Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
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" J; _, ^2 L- C- X    Base case: fib(0) = 0, fib(1) = 1( ^5 C  _& Q: F# F% g
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    Recursive case: fib(n) = fib(n-1) + fib(n-2)- C; H% s3 s8 Q( V

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def fibonacci(n):
, ?& E) E3 W7 W" Y* y7 j, S; a% Q1 }    # Base cases% A1 e" Q. S+ d' ]) d0 X9 o: i
    if n == 0:6 {' F- e0 m$ B' ?7 x
        return 0
  M. J8 B" q. H' g. ?6 I; e    elif n == 1:
9 [7 q1 G6 Z/ U" V8 Q9 M        return 1, c$ q* G; T6 _: [7 ]
    # Recursive case
: }0 K/ {. |7 g8 q    else:) @+ q) k/ U) Y6 q
        return fibonacci(n - 1) + fibonacci(n - 2)
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0 R1 N( u2 Y3 x+ j% c# Example usage% q2 j; @1 u! k. @& \
print(fibonacci(6))  # Output: 8
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Tail Recursion
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7 l  u3 `+ u$ ?$ {3 f5 `  c! d" bTail 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).2 Z- V- E. W2 ^5 Q2 A' j# b
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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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