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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& a3 K% M! z! i) J( U( i

& d: \/ w& w5 q0 A& g( D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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& p$ J  |. R; \7 \! r. Q" y 关键要素# L7 w9 o4 G4 q
1. **基线条件(Base Case)**
/ e/ g! ~, R9 a4 K# u6 D   - 递归终止的条件,防止无限循环( o* h" u, ?# P' x" Y( r. x1 ]
   - 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)**2 K% |+ k  f- V4 P- d
   - 将原问题分解为更小的子问题$ {( F; d8 V% l1 M
   - 例如:n! = n × (n-1)!5 V$ n0 [% B( ]8 m1 g1 w& y
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经典示例:计算阶乘$ m/ b: Q% H# \, P) ^' y
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def factorial(n):7 D* @! v* J+ T$ U" r( s
    if n == 0:        # 基线条件
; a( T+ y. @! f* c        return 1
4 O4 U" Y3 X% i0 h9 e7 |    else:             # 递归条件' i- V5 r8 b( ^7 E$ a
        return n * factorial(n-1)4 [+ x1 p1 m+ _6 E! w1 d9 _/ g# C
执行过程(以计算 3! 为例):
4 g0 X; j1 r1 M  h" _/ B4 Sfactorial(3)' i4 i3 G. _0 X7 H' u
3 * factorial(2)2 R* J$ _0 X1 u4 z" C( C: F
3 * (2 * factorial(1))% D! M3 I& J1 y/ K1 x0 H' `
3 * (2 * (1 * factorial(0)))5 r. E2 M/ d1 |3 J
3 * (2 * (1 * 1)) = 6
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递归思维要点. j' y! c  I" k
1. **信任递归**:假设子问题已经解决,专注当前层逻辑
8 @3 c3 p' k' i% e( R2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
" U& h8 f, N' E# R% d3. **递推过程**:不断向下分解问题(递)
: o* w" a/ ~# _" d4. **回溯过程**:组合子问题结果返回(归)
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9 z8 c0 V9 }, u% U/ e; ?. a) _ 注意事项3 G: @8 ?7 L. }8 y6 A  {4 C
必须要有终止条件
. M4 `1 x- \! h# M) t( R, `/ f递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
4 b1 @# a, n, w& L9 T! M某些问题用递归更直观(如树遍历),但效率可能不如迭代
$ s6 m; b( h- J  d; j1 h1 G尾递归优化可以提升效率(但Python不支持)
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递归 vs 迭代; \  L+ z/ C+ P+ C' x3 n% O! C# ~8 P2 }
|          | 递归                          | 迭代               |9 E2 o+ h0 z+ O% n' K: |
|----------|-----------------------------|------------------|
6 z! d/ D# K5 e$ h| 实现方式    | 函数自调用                        | 循环结构            |
2 C/ {* p) {4 L' Y( _+ @* h| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
0 O9 s' b+ k: j$ X, ]7 M| 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         |: [. a: W( U1 J0 ]' x! y. \
| 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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5 U. V/ P6 `$ f, v 经典递归应用场景
2 B- A% t7 i. B- T0 n4 T1. 文件系统遍历(目录树结构)
% W9 w9 z: Q, D( q5 S+ v4 y2 E2. 快速排序/归并排序算法# y4 |0 ?& l) V6 G+ Z( g5 m" q
3. 汉诺塔问题
* J, P' t% l! q+ c4. 二叉树遍历(前序/中序/后序)# K6 n  t! \" z% k
5. 生成所有可能的组合(回溯算法)
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9 d9 Q8 B4 q! v( q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,
( q" K8 t6 _( @& d我推理机的核心算法应该是二叉树遍历的变种。. f; \& x* q7 [* j( F4 a
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:
0 r5 f' V& N9 H( }/ e$ [Key Idea of Recursion
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A recursive function solves a problem by:, j/ X: S+ k0 ^7 o( b9 R2 d

3 b# b" |' u" T- [: v) Q6 ]    Breaking the problem into smaller instances of the same problem.1 Y/ n3 T$ K3 _* p1 b8 }% f* s# D5 U
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    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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Components of a Recursive Function
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& _  u9 g- }/ L" u    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.5 C3 `- Q: y& l+ o$ ]# p. P/ P
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        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.
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, N1 O* T- x! l9 j& X    Recursive Case:! R' `% ]2 G: V# ^
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        This is where the function calls itself with a smaller or simpler version of the problem.9 o7 U) O$ a- _

' H3 A% c0 E5 w& [9 @* q9 U        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).: O6 p3 h2 w& Y7 W& L+ ]" x' |
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Example: Factorial Calculation) m; l  v# I: g! B
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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:, P# X, D6 X/ Z  U0 V9 V

* ^) i; E+ K8 a    Base case: 0! = 1
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    Recursive case: n! = n * (n-1)!
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& |1 a; M9 h) S& HHere’s how it looks in code (Python):
3 I8 N+ v+ R7 w. Y# Spython
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def factorial(n):
" y$ V9 X0 l) Z  C$ B) h    # Base case/ R3 q: Z7 Y% ^! J  v6 _; e9 y" P
    if n == 0:
$ t% M* e2 m5 U/ d5 T0 \+ q" S        return 1- t; B& }6 @: N3 A( ?; D
    # Recursive case" ?! b7 q- M) A: {$ ]/ x
    else:
6 ?; f! t7 d0 P; d: ?7 n: R4 t! P        return n * factorial(n - 1): ~5 {: Z# H8 F; {2 P" V

3 z  p* O( q4 y" w& X* y6 O, T# Example usage
/ j3 [8 x; T# l" `print(factorial(5))  # Output: 120( j/ m  x& a- `5 C

, \! r  F" x) K4 HHow Recursion Works) N8 N+ T: s4 J( M5 G1 Q
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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.$ h) {8 e* ^- X% K' o$ ?+ A
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    These returned values are combined to produce the final result.
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' v, `" o4 O% f: G3 p, ^5 DFor factorial(5):
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! `* X( b+ Y" l  A- J4 Gfactorial(5) = 5 * factorial(4)
! ^# [* ^) r" g/ afactorial(4) = 4 * factorial(3)
& {. c4 c! t, k& w. l$ Qfactorial(3) = 3 * factorial(2)* C- s$ K) I  v6 U2 S" f
factorial(2) = 2 * factorial(1)
4 M$ l- s8 T" q# c. d, c' Ffactorial(1) = 1 * factorial(0)( @( Z) E; X! i- ?1 L" ^
factorial(0) = 1  # Base case
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Then, the results are combined:# Y) d% U7 r4 {0 a4 g" J

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factorial(1) = 1 * 1 = 1
$ y) s7 z; Y& b" [factorial(2) = 2 * 1 = 2
9 A- |$ g# s8 [  xfactorial(3) = 3 * 2 = 6. A" ], t: f# J" y& ^7 l, K0 J9 D% M& M
factorial(4) = 4 * 6 = 24. t$ ?' t9 }) s  {" H- ]
factorial(5) = 5 * 24 = 120
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Advantages of Recursion
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, l, |* F6 M( `9 ?- c    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).( r. p( n1 ~$ S/ `) B5 o
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    Readability: Recursive code can be more readable and concise compared to iterative solutions.
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: ~& P3 Y  S  b1 b% r  yDisadvantages of Recursion8 p4 t2 R7 U% O6 ?# c
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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.8 ?" ^) \2 b* d/ D, j

* Y7 }( p# T) _9 D    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
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When to Use Recursion
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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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/ |2 O2 X6 T/ z1 S8 E7 w    Problems with a clear base case and recursive case.
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) y' U3 k. ~3 T' [3 g0 g+ EExample: Fibonacci Sequence
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The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:9 o( b% w' Q# Q- \6 I
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    Base case: fib(0) = 0, fib(1) = 1/ H: Y- F$ M* A* S# f' U$ _
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    Recursive case: fib(n) = fib(n-1) + fib(n-2)7 u7 {7 E4 n+ C& N
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python
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def fibonacci(n):+ T& o- M# J/ b" c/ r: S6 M% ~  m
    # Base cases/ U/ q3 O  c2 r+ H5 c+ L+ {
    if n == 0:
1 n8 }6 ?# C' U9 u- q        return 0
3 V& D- e: M5 P1 `, \    elif n == 1:
) x3 W. e/ x3 S. P! U6 \        return 1
$ B3 m/ w2 S' _) W    # Recursive case8 y0 D6 q3 F# Z3 X/ n# u+ Z
    else:
* ?( R2 V3 {2 m; Q0 \0 ?6 ~' V        return fibonacci(n - 1) + fibonacci(n - 2)7 L3 N( h9 S# t8 W& n) z
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# Example usage
, Z9 \0 ^- l$ jprint(fibonacci(6))  # Output: 8
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Tail Recursion! J; d7 x9 e. s1 D" j1 `; q

' n2 X% A. ^" \+ g7 w, @9 {: GTail 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).% |+ ^6 q+ C3 u3 a& x3 f7 J

! X8 g" ^9 t9 `+ V' @) }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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