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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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* b) S: ^: ^/ U8 k& f$ l' [解释的不错2 @4 H2 p1 M( _/ |
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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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' E% n; {: Z; l5 @ 关键要素  P0 @, i1 Y$ u. j8 p
1. **基线条件(Base Case)**
  d/ a" l9 a0 a7 o   - 递归终止的条件,防止无限循环2 X- e! r" f% [1 |6 [6 o
   - 例如:计算阶乘时 n == 0 或 n == 1 时返回 1" y6 w, W6 [5 W& ]

. `% e( m: V7 N6 Z; Q* W+ U, _& L, D2. **递归条件(Recursive Case)**
- B0 J  z4 o& r5 \: T   - 将原问题分解为更小的子问题, M. f2 O0 `' v0 i
   - 例如:n! = n × (n-1)!" J! \: H( f2 X. S

& |  g1 k! I6 X' A7 }3 d/ Y 经典示例:计算阶乘
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def factorial(n):
0 @5 K4 L, `! ?    if n == 0:        # 基线条件
6 U3 t2 s( _: `+ f& `3 Q        return 1
6 h$ F, R; X' \/ U    else:             # 递归条件$ T( |& u1 D& ]4 p8 z, z
        return n * factorial(n-1)% Y3 `. o) v- j, B- {8 ?
执行过程(以计算 3! 为例):; i+ N' x4 E& i% E; N
factorial(3)
' b# K" K5 B3 g1 V3 * factorial(2)
! X8 V0 e0 `2 o3 * (2 * factorial(1))# r. R9 T1 V6 [" m
3 * (2 * (1 * factorial(0)))
0 T! g" _6 s3 I6 _- o& _7 T% S3 * (2 * (1 * 1)) = 6) M2 H' p2 m  c" R, Q+ k

" v$ c8 R, `6 g 递归思维要点- {! Q5 L$ }8 t* c; {# o) |
1. **信任递归**:假设子问题已经解决,专注当前层逻辑
5 e$ t* q+ C6 Q# K# x3 c6 ~2. **栈结构**:每次调用都会创建新的栈帧(内存空间)
3 Z/ S- T, l5 v3. **递推过程**:不断向下分解问题(递)
9 B& u; J6 m- Z( T' F; x4. **回溯过程**:组合子问题结果返回(归)9 Y4 M6 D0 b+ e5 `" m( X6 U; }% a

$ n9 D1 |7 {, D1 d9 j 注意事项
, `! B+ F4 V* `+ T7 h% {4 T必须要有终止条件7 l8 ~# n- v. y- X# ]
递归深度过大可能导致栈溢出(Python默认递归深度约1000层)8 @( t% D# v! p
某些问题用递归更直观(如树遍历),但效率可能不如迭代
+ @* e; S% e6 i6 H4 f& f尾递归优化可以提升效率(但Python不支持)2 ^9 U! q7 {" Y
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递归 vs 迭代+ M$ I; L: v  j- Y8 C. n" G
|          | 递归                          | 迭代               |* V) C  q8 s8 {
|----------|-----------------------------|------------------|; ]' n+ M& a& N, r: ?, g, ^
| 实现方式    | 函数自调用                        | 循环结构            |
* o5 N" Q( |% H- a" q( H  N/ [| 内存消耗    | 需要维护调用栈(可能溢出)               | 通常更节省内存         |
2 `4 x- ^7 {) l; F9 c) e| 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         |
) N, c1 ~: g% }3 s) V0 E$ f| 适用场景    | 树结构、分治算法、回溯问题等               | 简单重复操作          |/ o! `/ b& K6 t1 n+ t! y
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经典递归应用场景
5 q) T0 X: a( C7 q( o) S7 K7 T1. 文件系统遍历(目录树结构)
2 l8 q3 O2 s# b' t# d# U2. 快速排序/归并排序算法
9 l1 g* Z9 V/ U8 k! @& [6 {" }3. 汉诺塔问题
+ t2 V8 H1 O7 P$ r9 A; |4. 二叉树遍历(前序/中序/后序)4 m; }6 A& l! S/ }  e& R, d
5. 生成所有可能的组合(回溯算法)
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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,3 r  K8 }: ~6 e9 h3 O
我推理机的核心算法应该是二叉树遍历的变种。/ E; i8 M- u: W: @
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:
% x) D; w1 r( `1 p* DKey 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.9 Q( F9 D, a1 I  B. Y
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    Solving the smallest instance directly (base case).) P( J( [* M+ b: I# D5 \

9 `0 O( m' `' P0 Y1 E    Combining the results of smaller instances to solve the larger problem.6 x9 F: u. a2 f
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Components of a Recursive Function: \3 X0 a" a" M- s  c) e, {
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    Base Case:" G: \; O/ x& B* Q3 Q3 O3 u

. u; k; ^6 d& m& A% o3 l; ]        This is the simplest, smallest instance of the problem that can be solved directly without further recursion.' D8 u' @! s& u( J
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        It acts as the stopping condition to prevent infinite recursion.
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: z) J; U6 t1 n! E; H  x        Example: In calculating the factorial of a number, the base case is factorial(0) = 1.& I, I/ Y& F) L! i0 x: g0 a

+ R+ V- [- N5 U" i' T    Recursive Case:
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        This is where the function calls itself with a smaller or simpler version of the problem.$ X8 t6 o5 `* A! n6 |8 J4 Y
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        Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1).
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: g* t: W; i) ^+ B- pExample: Factorial Calculation7 P/ t/ X! i; ]# S. P8 o, R
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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:, ^; B1 N+ n) N  w
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    Base case: 0! = 1
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    Recursive case: n! = n * (n-1)!
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Here’s how it looks in code (Python):; D5 g% S( D$ M9 }6 O
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def factorial(n):  N( G1 h  f: J. I! @7 j4 ~
    # Base case: d6 g! w2 D1 I/ c" ~  B  C" R9 r& F
    if n == 0:* ~% s' l  @/ z9 D/ C
        return 1
6 H; X1 A4 U, A; V% h: t    # Recursive case2 {' }+ H3 t$ a3 n$ [3 b3 {: Q/ ^, m
    else:
1 h: O- H4 s  q' T9 |        return n * factorial(n - 1)) j4 {3 w% Y$ K9 e

* [5 J/ J" Z  Q9 l7 G: A% Q, C, L# Example usage
& S- ]5 r# J& o8 fprint(factorial(5))  # Output: 120
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' U- c" Y; ^* }How Recursion Works  Y* D0 u& K" a  X

- w5 F/ q1 d5 T  W$ a    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.3 E  c0 A. P% V4 N, a8 N

, v7 U5 `( o0 P: d( O4 O    These returned values are combined to produce the final result.
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7 u) A" y) v4 D" J' b7 _For factorial(5):
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factorial(5) = 5 * factorial(4)* n3 G0 s+ I5 R' w  t
factorial(4) = 4 * factorial(3)( {# e2 C8 v- r6 [9 z% N! Y) q
factorial(3) = 3 * factorial(2)1 {3 [4 M! G, B2 W
factorial(2) = 2 * factorial(1)
- r6 f. |5 N! k! O7 T0 J8 ifactorial(1) = 1 * factorial(0)
3 i/ a% Q, A: t! y% p8 D1 Jfactorial(0) = 1  # Base case  [: c' g/ |1 o" a& F5 D
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Then, the results are combined:
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factorial(1) = 1 * 1 = 1) t. h6 {3 C- Y$ U, N
factorial(2) = 2 * 1 = 2  Z& z. q( s0 l
factorial(3) = 3 * 2 = 6' R, m! `) O8 `6 ?; ]' k( ~
factorial(4) = 4 * 6 = 24$ F/ k, g' F& R  n
factorial(5) = 5 * 24 = 120
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% F/ H" |% \3 ]Advantages of Recursion
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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.7 M1 @3 B3 S" R+ J+ {  l
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Disadvantages of Recursion( o; Y* W9 [: _% C: s+ t
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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).8 K+ l3 d% q/ f6 M3 T: ]; G
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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).' i0 `, E8 u7 y& h1 M# q0 S

9 b3 v  r" x7 E8 y    Problems with a clear base case and recursive case.
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4 l9 ]7 V2 V" p3 W% d6 Q% b# S. ZExample: 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:% x8 }. j5 I# k% t7 p6 N- E& b
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    Base case: fib(0) = 0, fib(1) = 1
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6 O% v( F; }% b$ x1 _% }    Recursive case: fib(n) = fib(n-1) + fib(n-2)
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python
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4 s: I+ n5 S' x7 F; d" \def fibonacci(n):
6 U+ U0 q8 D6 d6 n. o2 E* W    # Base cases. U: x% u8 C: k) w# t
    if n == 0:
; a: u6 s) |$ p  C        return 0
4 T9 \4 @4 Z5 N. R# \" @6 T    elif n == 1:
' Y9 T1 m- p- E* @; B( {( k        return 1' ]. A& }0 T3 P) |/ F9 p  J
    # Recursive case6 ^+ M/ E- N1 l5 {
    else:0 z, H5 D- M! w# P" B
        return fibonacci(n - 1) + fibonacci(n - 2)2 }% Q$ p% }+ U3 k7 D

3 S9 a( E4 ^9 M+ w# Example usage; ?& P# t3 \  Z2 s1 y
print(fibonacci(6))  # Output: 8( h$ H) R& s  X! }1 [

1 E4 T6 x8 j( e* xTail Recursion  w: B; g6 t) b' t4 t- T

/ U/ D, B- D# K4 I$ O4 F5 H1 DTail 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).9 E, h3 c, J1 K4 ]1 i7 @/ v1 ~9 _

1 C: y% Z8 ]3 p8 S5 b2 JIn 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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