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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 1 _4 C# g" i; ~, P+ r  t! S
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解释的不错
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4 |( w6 C2 b8 {: x4 Q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。' q4 h4 o0 s$ S* i9 G$ y

/ m/ w# r: F! U1 | 关键要素! {3 i; o6 u6 o# O2 T3 f; v  a
1. **基线条件(Base Case)**" t$ i8 \  P' A' @' ], f- b
   - 递归终止的条件,防止无限循环
; ^1 p% r( K4 Y! u: @5 B   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1# m2 [# h3 F9 f2 h* f

* b5 R9 B, B. q  q- G, r6 M: V6 L' k2. **递归条件(Recursive Case)**
# F7 V0 b( o- U& }. P   - 将原问题分解为更小的子问题
9 e+ ~9 S2 o1 X. f3 E- W   - 例如:n! = n × (n-1)!+ p% Y' H3 C3 }# u8 J( E
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经典示例:计算阶乘% f9 c' B" B8 b
python
8 V% V# m* L# A( E, A+ i; ^def factorial(n):* T7 z; ?  u" c3 G
    if n == 0:        # 基线条件
% A8 N7 w1 @4 g, q; ^2 n( N: F        return 1
# v5 W9 e1 z  K3 U4 M    else:             # 递归条件/ p: z4 Y8 ]$ w" ]
        return n * factorial(n-1)3 R' E. M9 S( d# P# E9 f
执行过程(以计算 3! 为例):5 Z) q7 f4 w, d) u
factorial(3)# e$ L7 b& z; _& {" [
3 * factorial(2)
5 s2 O* X2 w. d2 y  L0 F6 _8 B$ d3 * (2 * factorial(1))
8 U4 L; u9 [  l; N& E$ k3 * (2 * (1 * factorial(0)))  b0 X* `% z8 m+ y
3 * (2 * (1 * 1)) = 6
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4 w1 Z% P# b: v/ q+ K& h 递归思维要点
7 N2 Z  O% W7 T( d1. **信任递归**:假设子问题已经解决,专注当前层逻辑6 N8 G$ W" c) a: y4 E
2. **栈结构**:每次调用都会创建新的栈帧(内存空间), L5 E8 _  E0 v; Q4 i8 o
3. **递推过程**:不断向下分解问题(递), W/ G" x' Y1 D& P7 _$ q( B" M& T8 f
4. **回溯过程**:组合子问题结果返回(归)" i, W3 S4 t0 s( c6 u6 H' c
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注意事项( R4 y3 v8 g$ R2 z" i
必须要有终止条件
8 l, M8 n! U, V4 s递归深度过大可能导致栈溢出(Python默认递归深度约1000层); z3 n1 V/ I7 A; a! Q
某些问题用递归更直观(如树遍历),但效率可能不如迭代
2 ]0 ~# k* R- R  I尾递归优化可以提升效率(但Python不支持)
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! S- b, o2 G$ W 递归 vs 迭代9 {& d3 @! T  Z5 \$ v" T
|          | 递归                          | 迭代               |( ]: B6 ~, x0 [: A+ ~
|----------|-----------------------------|------------------|
5 T) c0 E1 _: L! C; [, j7 @7 W| 实现方式    | 函数自调用                        | 循环结构            |
( T) \. l1 B3 T) m| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
5 n% a5 y# H; B+ || 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         |/ d6 b$ ^3 M2 G! u1 w1 v
| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |1 @) n$ _1 Q  ]2 N" b
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经典递归应用场景
" w, u. E% }3 @. V" h' m1. 文件系统遍历(目录树结构)3 `8 c% m! v2 w( J
2. 快速排序/归并排序算法
2 s/ n" H; X( H9 G$ a3. 汉诺塔问题
/ E* C4 h/ W3 q; L4. 二叉树遍历(前序/中序/后序)0 l# k+ V) B4 h% y; a
5. 生成所有可能的组合(回溯算法)! N, B% v- m$ W4 r
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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 P; _+ A7 Q) ~: G" W我推理机的核心算法应该是二叉树遍历的变种。. e* h- A0 r% z4 h6 P, R6 z9 e" h
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:3 Z$ Z  `+ F: u( B7 z% a/ a/ K8 v
Key Idea of Recursion
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A recursive function solves a problem by:
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    Breaking the problem into smaller instances of the same problem.
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    Solving the smallest instance directly (base case).- U' v: P+ q0 `. ~

2 Z* C) ^; Q+ S: D9 l& I) f$ X    Combining the results of smaller instances to solve the larger problem.7 M$ R1 T4 S3 [% @9 L1 W6 o

' [, i; t3 k# ?: ?7 t: i% Y1 JComponents of a Recursive Function
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  `2 D* X) m* e7 |# O( Y0 U0 i    Base Case:5 N9 s; C% S! U. w; w7 U( a

& m4 e- j' H6 X        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.- y  r0 B, t+ K
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        It acts as the stopping condition to prevent infinite recursion.7 q  Z* C6 Y; O/ b) s( ]
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        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.0 c1 S# q  s5 S, a2 Z4 V5 `# v
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    Recursive Case:
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( Z% N; z% H% u  o9 S' {( A        This is where the function calls itself with a smaller or simpler version of the problem.
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2 ^/ |9 v* h( `# u        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).
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/ Q- j; G5 g8 C: L- mExample: Factorial Calculation
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% [0 }" K6 x* @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:9 \0 h$ s% G5 l% P" w& K9 L

# _. t3 H0 s6 r7 q) J    Base case: 0! = 1, k1 O7 A+ Y( I( S0 L3 x! H

4 i  h4 ~0 K* J  Q    Recursive case: n! = n * (n-1)!
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Here’s how it looks in code (Python):
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7 I, l+ y3 u& H+ k  pdef factorial(n):
: H( w9 Q+ W' m) T' y" z    # Base case- |  ?/ P! k8 |
    if n == 0:
5 p3 [2 e1 X- ]8 N/ U& n        return 1$ s. f( Z$ ~. Q! t% @2 M
    # Recursive case
" B; ?* q- c/ o( g; j    else:+ @; @7 K8 B6 ?3 ?' q8 x- ]0 {0 T2 K
        return n * factorial(n - 1)
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# Example usage  z. n0 e6 {# k! p7 K# ?
print(factorial(5))  # Output: 120
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How Recursion Works: `1 a  C+ A0 X+ S. v% N2 T8 V. h
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    The function keeps calling itself with smaller inputs until it reaches the base case.
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3 z0 n% x. D/ j$ o" `% D4 a5 ~& G( m    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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$ ^0 f! K& G4 X4 r; T: WFor factorial(5):7 s( P1 K4 H. S' U

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factorial(5) = 5 * factorial(4)
: Q! r7 Z: `( c) G0 vfactorial(4) = 4 * factorial(3)! h" {( V! h# {6 M5 u
factorial(3) = 3 * factorial(2)
+ i/ P3 L) [: b2 D4 s) {0 Lfactorial(2) = 2 * factorial(1)
  B! n; F" m: s: j) L/ v1 ffactorial(1) = 1 * factorial(0). R+ X" A4 A' C, x5 w  Y
factorial(0) = 1  # Base case0 g3 ~7 H$ `: g7 W  ^
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Then, the results are combined:0 R5 \! m) ?7 z  \" B+ e2 j0 T0 i
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+ h4 o- a( p, ifactorial(1) = 1 * 1 = 1
9 f9 ]5 G) n% j- |9 ufactorial(2) = 2 * 1 = 2) c5 R9 x% v! }5 e4 k2 L5 ]
factorial(3) = 3 * 2 = 6
( }9 [8 A/ o4 Z" N- _) cfactorial(4) = 4 * 6 = 24
' p  q- O7 ]7 n4 T$ @factorial(5) = 5 * 24 = 120
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0 |9 m2 L; J; q8 a1 NAdvantages of Recursion, d* Y2 a. B. K0 r0 w5 f
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    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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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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    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
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! E% O& S! a& p/ p& N" [: j' UWhen to Use Recursion* K% n; s6 i: z
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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.4 d" }" ^% ~6 s4 L3 V
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Example: Fibonacci Sequence. U3 f9 K3 e5 b/ V6 \

  ~+ h* C) H5 i5 Q" m  E/ gThe Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
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    Base case: fib(0) = 0, fib(1) = 1
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4 W0 `) F% E" f) A    Recursive case: fib(n) = fib(n-1) + fib(n-2)
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python
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& s; [) D1 c, F$ P0 b1 rdef fibonacci(n):+ N/ l4 u+ W; p! M7 n# K+ X/ r+ h# t7 N
    # Base cases* w6 p* q# j: h
    if n == 0:' L; i5 D* e( d4 x/ P- ^3 \% x* I) P% O4 H
        return 01 K( A4 ?& b* w6 w0 a8 a4 r
    elif n == 1:
& V6 x9 L' F8 F3 v        return 12 l+ x4 j5 _+ d/ X. r
    # Recursive case
* S6 M3 i* A- Q; m    else:
, }6 R5 n0 f$ T9 P) o        return fibonacci(n - 1) + fibonacci(n - 2)
# ^2 m1 d$ q9 N! \- K4 _, B& G$ {: I. {. F1 @4 y+ l
# Example usage# w% A  n& X( `% F6 M+ w
print(fibonacci(6))  # Output: 8
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Tail Recursion. e. g2 w* w8 {

0 R% R! s( K, m4 P/ ITail 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).; L9 H8 x! i0 ?2 W# Y% f
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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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