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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6 q  V8 Z0 G8 i+ x4 v) f& N. w* S
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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1 ~, Z2 e& Q  [5 f
1. **基线条件(Base Case)**
6 D+ n1 {( ?: L# [   - 递归终止的条件,防止无限循环- C; t$ o7 O, l
   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1
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& E3 Y9 \# P9 p- d2. **递归条件(Recursive Case)**- q/ ~) N  S! j+ x. c. x6 q; p
   - 将原问题分解为更小的子问题( u$ P4 G1 S! Q* d* G
   - 例如:n! = n × (n-1)!3 b+ N' K+ u. L5 k! |- N

2 G2 z4 S; N2 ^& Q( D4 L* Q 经典示例:计算阶乘6 ^* U: ]3 t" y/ O
python9 Z$ e: u1 f# `, W
def factorial(n):9 |; }3 [$ x. |, {
    if n == 0:        # 基线条件, r! y4 s9 O0 b7 E! [# h
        return 1; a. {5 d2 H& O; ?& J+ {
    else:             # 递归条件
* K  g9 u9 _0 `7 f; w' i: H$ W1 d/ k        return n * factorial(n-1)
$ F7 q$ ?4 _. F) z/ I# X  A执行过程(以计算 3! 为例):" _9 `6 p+ R) a+ q, h
factorial(3)
5 c8 l2 u; ?# z. I3 * factorial(2)
  |1 u/ N6 Z5 I' D3 * (2 * factorial(1))9 W( o- v5 g: l* E$ Q
3 * (2 * (1 * factorial(0)))
8 q( P* e& l; ]. r2 _! h3 * (2 * (1 * 1)) = 6& F4 a% L! f! N

+ }- k2 _+ F: k  V, x4 N 递归思维要点
, l# s# ~: q) x4 @# q. ~1. **信任递归**:假设子问题已经解决,专注当前层逻辑/ n, w& `. O7 {; X& E
2. **栈结构**:每次调用都会创建新的栈帧(内存空间)# j" ]" T6 C# g! ~1 X
3. **递推过程**:不断向下分解问题(递), h$ e6 c- \6 @" P! S
4. **回溯过程**:组合子问题结果返回(归)
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注意事项
6 P9 d" n! o  \  l" R% v0 i% v3 ]* n必须要有终止条件4 z, D$ W! @7 V' f$ h* b
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)
+ R4 S% G, G4 v0 f某些问题用递归更直观(如树遍历),但效率可能不如迭代. l+ c6 w9 @2 O  @
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: H+ K& K4 @  w! |- p" \
|          | 递归                          | 迭代               |
. x" l' Q7 g! a% q9 ~- _|----------|-----------------------------|------------------|. n# J( f7 y5 b3 s
| 实现方式    | 函数自调用                        | 循环结构            |
. e# R  D0 _8 i: h| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |4 G2 ~2 s0 @( a
| 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         |
- b7 S+ w; n2 ~5 Z! c. b) L| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |) f6 z8 _$ ]  j, p

4 b0 l+ C( i7 Y" N 经典递归应用场景1 j2 M- f# H, d( e( O. I; {
1. 文件系统遍历(目录树结构)( x% i3 R4 h8 k3 J
2. 快速排序/归并排序算法: @8 M+ [2 d' a6 T3 w* k
3. 汉诺塔问题1 s5 L0 Q0 j2 M8 z3 g/ f; E3 M
4. 二叉树遍历(前序/中序/后序)
. r6 n$ @& h/ X; }2 y5. 生成所有可能的组合(回溯算法)
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& |% g% B; S' w; X7 M) b9 u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,/ Y/ ~9 j3 i' K% k& q0 k
我推理机的核心算法应该是二叉树遍历的变种。
( g& I7 V& A$ |1 }% R6 @- ]  s1 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:- g% S1 n" j: I; c) [) i! A* ?
Key Idea of Recursion, X! D2 o' J' @# G; O' X* C* [' V
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A recursive function solves a problem by:) u9 v$ e6 n  S/ v6 I$ t
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    Breaking the problem into smaller instances of the same problem.
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* e; ]6 u& P, A3 {3 Q    Solving the smallest instance directly (base case).& s5 D% ?" p( G( m6 ]
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    Combining the results of smaller instances to solve the larger problem./ H, y$ d4 U: O) |/ }& y4 ~
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Components of a Recursive Function
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) z. p' N0 R  ?2 L/ E    Base Case:
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7 J" L$ v/ X  i- D  Z% a        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.
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) y: d  f, R8 ?) r; H        It acts as the stopping condition to prevent infinite recursion.
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  s5 l3 o5 v' m$ P( L& Z        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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. q  K0 c: l: y" h        This is where the function calls itself with a smaller or simpler version of the problem.: r- v! ?8 C0 q# b

7 C* S0 H' P& H" }        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).1 U% b% i8 x1 z' k
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Example: Factorial Calculation
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+ u; r- T4 f8 t7 D' v8 a( k9 QThe 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:3 V% ~. a. `$ t/ L

2 |1 J7 s% @8 S    Base case: 0! = 1$ T" k" T) A! A3 q! S% X

3 N, b! s0 g' s* `1 W5 w    Recursive case: n! = n * (n-1)!6 M) t1 y: X# Y$ \+ ]

# ?/ w+ K4 m* x7 d/ [Here’s how it looks in code (Python):
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) E0 d# u8 l, M% ]& jdef factorial(n):
, a+ @0 s# Y; e+ l- Q9 c) {/ b    # Base case
. y( ^/ q8 D/ x8 Z( l; M/ V    if n == 0:
9 N- S! Q# Q8 e# [$ S) k        return 11 ~' i  @) O  C: N
    # Recursive case
4 `! m2 w! Y# V    else:
: r# Q- D- h' c- X( t( @& [9 L        return n * factorial(n - 1)
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# Example usage, b. e4 n$ A0 A, E. v. Q& {* Y
print(factorial(5))  # Output: 120/ v/ ]% X9 \- T

; z6 b3 b' c4 S' e4 x5 j4 ?7 kHow Recursion Works: r. m6 o3 F7 |
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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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) l6 U! d* c) H$ u0 s! L    These returned values are combined to produce the final result.
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For factorial(5):
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$ R0 }* Z0 k# y% L# u1 e: hfactorial(5) = 5 * factorial(4)/ K8 l7 ~& g7 A
factorial(4) = 4 * factorial(3)7 K& \( a/ Q. M5 T
factorial(3) = 3 * factorial(2)1 a0 y4 B+ k  K* F2 m, g7 ^' e/ r
factorial(2) = 2 * factorial(1)
0 y4 h4 _  U# U, A% c# Qfactorial(1) = 1 * factorial(0)5 X( o! C+ F! p3 C* \" c) ~# I) ~
factorial(0) = 1  # Base case
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Then, the results are combined:/ h: w" r* Y) M& i9 K
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factorial(1) = 1 * 1 = 1
7 \9 ^. x; B: L( R" Zfactorial(2) = 2 * 1 = 2
% L, A, o# A% F) j5 ffactorial(3) = 3 * 2 = 6
& v; }' Y: d& Sfactorial(4) = 4 * 6 = 240 p7 a' Q* i/ }, f, H1 x9 t! O
factorial(5) = 5 * 24 = 120. o3 i( a" q; }

7 t+ u% |# r- KAdvantages of Recursion
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4 }; M* F) ]/ v: V    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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$ U9 A1 U& Z* m/ H# U    Readability: Recursive code can be more readable and concise compared to iterative solutions.9 ^1 O; E$ o$ _7 P, [7 J
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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.4 j3 j" T* ~1 M# b( P" v, B
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    Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).! a5 f+ E' i8 e( ]7 v& D  U" b. D/ H/ R

( d4 I1 M! ^6 z' f! r3 |When to Use Recursion6 j/ Q" g! q3 o/ \, Z8 y! Y

9 t, u7 B3 I3 S    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.; [: P" W# ~/ r, a! j7 E; {& J* H
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Example: Fibonacci Sequence7 \) E1 v3 W: x5 q: x( @

+ _3 {6 J! }, v% z4 s9 vThe Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:
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5 l+ @# j: \0 M& p+ ]    Base case: fib(0) = 0, fib(1) = 1
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    Recursive case: fib(n) = fib(n-1) + fib(n-2)- b6 e/ h/ j8 I4 M- {2 L
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python
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& K6 d* G: h# ?$ J+ [9 |+ jdef fibonacci(n):, O6 B8 a/ U# S9 D' @) Y
    # Base cases
6 V3 [$ t' e" P2 [) ^    if n == 0:
& W% I5 @- E' H, q' s+ X        return 0
/ H) ^% Z) H$ n8 b    elif n == 1:, P, ^: S7 x: u2 H# \* D
        return 1. O# J3 _7 q; R% U+ T
    # Recursive case3 L8 t' u9 \; k/ A9 e4 }) n  P7 U
    else:
0 t# s6 T" z6 R& o        return fibonacci(n - 1) + fibonacci(n - 2): p  E) F; H& N
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# Example usage2 Y; E" V! Z: B( E$ w
print(fibonacci(6))  # Output: 8( }  O7 R0 }0 W- O' s; e1 b

% l  ^' {+ K5 j4 A( Q. ~5 iTail Recursion( q6 B, n! [8 t* p
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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).7 ~: G* w" _! P

6 v8 I0 D% K. s* F4 Y* j* U7 ?% }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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