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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 7 D3 i( Z9 }4 R; O; X( I- F. S

! Q3 o8 s5 k' R+ }* v! q& V, |解释的不错
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7 g5 {( _2 \3 j, X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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% G6 x. P' m+ j$ m 关键要素
: ^* U) d6 P$ v3 v9 }9 x1. **基线条件(Base Case)**
- z5 N8 p% e) e" S) k6 v3 ~5 J7 t   - 递归终止的条件,防止无限循环
2 `0 f9 }9 m3 c5 M' |& V1 E   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1" ~. j+ \$ A. P$ ~" u5 s: {, Y6 ^

$ V6 S& _; ~  O/ \2. **递归条件(Recursive Case)**  R) K% f5 @4 |0 A& X5 R
   - 将原问题分解为更小的子问题' S, u$ v' u  p5 j* l9 S
   - 例如:n! = n × (n-1)!: E9 z4 B9 E- H2 z$ b) o

1 q/ ?$ l6 j# I! E3 J+ K3 l" d  U 经典示例:计算阶乘3 G1 O8 e3 G5 V
python+ o# \2 E) M! `2 O+ Z& |. U0 M
def factorial(n):' _9 |' s& @2 p  u
    if n == 0:        # 基线条件2 f7 v4 \* C0 {( M6 d4 }& @
        return 1
* x4 t# S8 N6 Q3 n  R    else:             # 递归条件
- J4 C/ X7 h6 ?4 K: ^( m- ^4 H, l1 ]        return n * factorial(n-1)
8 J3 H2 @0 e$ Q1 Q执行过程(以计算 3! 为例):( k3 o* z  y% W* m5 `# n5 \% p, K# n
factorial(3)
; `8 B3 N6 ?( L, M% m' W3 * factorial(2)5 N# T/ H" [, h, r
3 * (2 * factorial(1))% e* H) ?% V6 Y0 @
3 * (2 * (1 * factorial(0)))
( F2 F: E  O, h# T; Y! w* |3 * (2 * (1 * 1)) = 6
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递归思维要点8 u# {2 a; [0 N9 B! x1 r" u
1. **信任递归**:假设子问题已经解决,专注当前层逻辑
! z, i2 u) ^# h2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
! q6 e. d2 i4 o) ]$ ?3. **递推过程**:不断向下分解问题(递)
8 g8 t8 ?) [2 P4. **回溯过程**:组合子问题结果返回(归)6 E# g9 x  ~: R1 Q; e9 {
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注意事项$ e+ V9 {& ^: C( H8 |
必须要有终止条件
: a$ s- B8 @+ ]: y( C递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
9 e/ T! `) G+ ^7 R# v某些问题用递归更直观(如树遍历),但效率可能不如迭代
( V5 s8 p  I. l# p" |尾递归优化可以提升效率(但Python不支持)' K/ _# T% U' ~% y/ q: g% T
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递归 vs 迭代: n1 J0 a. F  V5 n  R
|          | 递归                          | 迭代               |
1 m9 j# ]" `5 f|----------|-----------------------------|------------------|
/ G& @& N' y) E+ E$ u+ g. B" }% ?( V| 实现方式    | 函数自调用                        | 循环结构            |2 C' S8 W( I* d7 C
| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |; S) }4 [9 j( A2 L, O# P$ P
| 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         |7 T7 k! _) Q4 w9 u8 `
| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |- U: T; r9 p3 g

3 Q6 v8 I& B/ s/ J 经典递归应用场景$ o7 s" W* l( u6 c# y
1. 文件系统遍历(目录树结构)
- J# ]3 @, o/ q2. 快速排序/归并排序算法+ m! w0 H: F6 r; N' i
3. 汉诺塔问题
2 S' ?' S/ ^2 J4. 二叉树遍历(前序/中序/后序)5 D! O: L' Y1 [
5. 生成所有可能的组合(回溯算法)7 A$ P/ [  {: F5 O+ j* s& g7 u* a

- h' }2 v9 p& s! v/ K' |; ^/ [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,. O: X3 o! O' ?  |9 F% D4 E; o" u
我推理机的核心算法应该是二叉树遍历的变种。
. f$ X7 Z$ T4 N# S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 V/ m% v2 H# j- I4 tKey Idea of Recursion) k& A% E2 M1 U4 m; g" d7 J
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A recursive function solves a problem by:! V- Q! q0 |0 Y  ~# c7 }# Z

" E, r% h- G$ P) H    Breaking the problem into smaller instances of the same problem.% ?7 a! N7 w) O! w4 x9 o; Y6 ]
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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' z9 z1 w/ \6 p2 _& w( z; q

# D0 y5 b6 s) H3 _* N4 `0 Z    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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( R4 E( G  j) d1 j" E        It acts as the stopping condition to prevent infinite recursion., `. S  a+ x) ]% i. x6 r+ U

/ ]/ |/ F: {9 F1 q* X  L6 }        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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9 g) b2 p) N7 R8 x) o* P/ F        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).! i) }( p$ K! q/ v2 ]& X2 q  \  z

, X3 S8 {/ o. c/ z1 d( R" |5 w8 q" ]Example: Factorial Calculation9 m/ h8 P- \# z" w1 m( s' D4 f
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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:& u4 A/ ?2 w) J' G6 j. b
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    Base case: 0! = 1
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# h3 D: e8 s& m) a7 e    Recursive case: n! = n * (n-1)!
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Here’s how it looks in code (Python):
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3 M( G6 A( O2 ]. k0 ~) h: Udef factorial(n):  a* R8 z7 c! Q$ W
    # Base case% p$ B. v# z% l6 a$ Q3 ?: X6 N
    if n == 0:
) ?' p( |6 k' K1 F, F        return 1# `/ w3 |) i$ ?: e) G: Z
    # Recursive case
: K, {  P+ E- n; U    else:
, r) I% J  N3 C        return n * factorial(n - 1)
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# Example usage/ s1 s% H0 G4 ~( P/ }' {7 a
print(factorial(5))  # Output: 120! T. f0 p9 N7 D0 p4 k
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How Recursion Works; E" J9 e) g; Z% P

! n" R" Q; A4 e$ B$ v0 E. m    The function keeps calling itself with smaller inputs until it reaches the base case.2 z: I1 A5 E/ {2 w' o
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    Once the base case is reached, the function starts returning values back up the call stack.6 N7 E" x7 D5 B# S5 J

1 w$ o8 n: I, M$ t  v    These returned values are combined to produce the final result.
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* b8 b1 {/ ~5 DFor factorial(5):
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3 a9 T6 \/ t3 Z# B1 {% Ffactorial(5) = 5 * factorial(4)( ^/ a; L+ K0 `3 g; {! V
factorial(4) = 4 * factorial(3)
$ Q8 l7 \  P% D1 S6 {& V% {factorial(3) = 3 * factorial(2)
# A- O0 g$ o+ `4 H$ jfactorial(2) = 2 * factorial(1)7 r3 N! Z& B' m8 u6 J* ^1 q
factorial(1) = 1 * factorial(0)
0 E7 B2 l& U* P5 L. Tfactorial(0) = 1  # Base case
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Then, the results are combined:
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factorial(1) = 1 * 1 = 1
/ S) w3 u' b/ F3 n4 ]% v8 R8 bfactorial(2) = 2 * 1 = 2
1 }; U" L0 D, Pfactorial(3) = 3 * 2 = 6. ~6 L, k1 [# _! e. v
factorial(4) = 4 * 6 = 241 a* F3 v" C* {1 X
factorial(5) = 5 * 24 = 120
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' H4 ~- }9 c/ {/ [( ~1 z( Y& KAdvantages of Recursion- S' Q, A% C9 r
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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.3 F, Q! }8 A% i: O2 b) h
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Disadvantages of Recursion
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5 L1 a) C& f# a; W; L3 V    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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. Z9 K# c3 Y. U2 n4 T# m    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).
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7 Y" M# N7 Z% J! k; h$ TWhen 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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    Problems with a clear base case and recursive case.
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Example: Fibonacci Sequence
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/ R3 |: Y- x1 L3 N4 R  ?The 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) = 14 I' t; {" W5 T: J7 }, T1 W; m6 \

( q7 q4 L! c" s    Recursive case: fib(n) = fib(n-1) + fib(n-2)
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+ K5 K: i; r, e, L$ jpython/ t* b+ G( O  f" m( v, m
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, N5 O% X$ ]) m4 pdef fibonacci(n):  o+ \- Y% W: p3 u5 P7 O) Y
    # Base cases& I6 k. l  I% G
    if n == 0:
) y  T+ z0 Y0 a  ?+ @        return 0/ j, u& B: ]$ X
    elif n == 1:0 c7 r' Y3 Y+ C( L6 f% d: [
        return 1
- Q* g5 G! i' U+ ?    # Recursive case
  F# P2 J6 x% g    else:3 J0 l) @6 I: T9 ^( W6 A! _: t
        return fibonacci(n - 1) + fibonacci(n - 2)! c- U* u% x( L- U

$ R% B( ]0 x- y4 a8 U# Example usage! [( _: o# n" Z9 |3 R
print(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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2 Q( n: N6 y6 z: l0 DIn 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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