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 % E" P/ M9 Q9 j7 }8 a
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解释的不错8 K M7 @- ~! Z1 D
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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。 0 v. V; ~+ j) ?, ? `* z& N8 l; ]4 @
关键要素! R5 ~2 b# ]& D% @
1. **基线条件(Base Case)** ; u$ v' k# w9 [* \, e0 r/ d - 递归终止的条件,防止无限循环5 c; o* n K, m4 X4 q5 U \
- 例如:计算阶乘时 n == 0 或 n == 1 时返回 1 1 h! _% i7 h! m& R. P3 w0 w6 {3 Q4 M , E) ~0 u2 s) w2. **递归条件(Recursive Case)**9 Q+ _# [" H2 U0 l, [8 m2 x
- 将原问题分解为更小的子问题 $ P) b9 T- g& W& t - 例如:n! = n × (n-1)!2 `- f" L( W6 e+ S" ?' V
8 J, a6 u1 }4 Y 经典示例:计算阶乘8 z4 O" n. n% K- A# J
python& O# I- g0 _6 @$ _9 y6 z. b' c6 T
def factorial(n): ( ~+ k8 c; X) [% v2 d9 F/ ]9 _) T if n == 0: # 基线条件 ( i* [! y8 J, ?$ r1 L1 i return 1 ) }6 t) L X; _2 y ^, D, y! w else: # 递归条件7 [5 ?2 n# h) C2 Y( Y, ]/ `' b( u2 P
return n * factorial(n-1)0 L" [7 [( V, x0 g5 G0 R
执行过程(以计算 3! 为例):2 \, Y# i5 O. r- R3 J4 D. x; d' Q
factorial(3) ) m( n: m1 I0 v: u& e5 v/ J' N8 f3 * factorial(2). `: ?5 _1 ?9 \
3 * (2 * factorial(1)) 8 E) r: }7 B6 R3 * (2 * (1 * factorial(0))) $ n: D0 D4 `) r% \ U. ^: F3 * (2 * (1 * 1)) = 6 3 ?& W; G6 R/ ]7 i: k( Z$ A6 ]: m/ g: S
递归思维要点, u3 V7 m: C% G! k) e+ o
1. **信任递归**:假设子问题已经解决,专注当前层逻辑 9 ]+ u" e! C! C/ V, ^; d' r2. **栈结构**:每次调用都会创建新的栈帧(内存空间)5 x/ Z9 c" x* X; T; P8 O
3. **递推过程**:不断向下分解问题(递) 0 U3 N; T9 h* O4. **回溯过程**:组合子问题结果返回(归)) L7 D f4 |( o
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注意事项# e$ y$ ?+ y8 n- y
必须要有终止条件) Q+ X6 L" ~" y' V3 `* @1 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 R0 @8 y8 i5 j' B4 ] j: b; M9 }某些问题用递归更直观(如树遍历),但效率可能不如迭代3 s0 D/ Y# \2 k# F5 V" k
尾递归优化可以提升效率(但Python不支持) # B: v3 D& h; m k* [ j. R7 r/ y3 e, e8 I$ ?! b3 E* O/ q/ @
递归 vs 迭代 o2 p x6 y" |1 \
| | 递归 | 迭代 | ' d2 s2 d8 P- Y1 f I- p: R6 V|----------|-----------------------------|------------------| 4 u& u2 M& d- X| 实现方式 | 函数自调用 | 循环结构 |( e9 L) @ ?4 Y7 Q* c
| 内存消耗 | 需要维护调用栈(可能溢出) | 通常更节省内存 |- Q- S: I. t7 _, R" S! G; _
| 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 | % l d2 Z% b7 M+ x, V o( {| 适用场景 | 树结构、分治算法、回溯问题等 | 简单重复操作 |, H& e G+ Y) e2 B. @
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经典递归应用场景 : X: x' Y' _0 e9 b |1. 文件系统遍历(目录树结构) . b* c, d* q1 X1 t- s: H1 a" f2. 快速排序/归并排序算法 R# X; k2 y% F$ R5 f0 t9 B" w* N3. 汉诺塔问题: v* D1 Q" s9 |+ S$ g) T3 O! x
4. 二叉树遍历(前序/中序/后序); S/ k, e5 X' c. H3 [
5. 生成所有可能的组合(回溯算法) , e$ b' z) n* \$ @5 R " V2 C0 N) \' Y, O) w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, 9 \/ b, p! x! P9 B我推理机的核心算法应该是二叉树遍历的变种。& E/ w% ~" f) p* K$ m6 R
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: + V8 @; v& N N& p0 ~& S- C; [Key Idea of Recursion 6 b7 y' j0 `( s8 ] B % W4 n; y2 X; L g+ d* G; IA recursive function solves a problem by:: o) A" W8 o# e; u+ L5 X: c
8 N( [( n7 e2 d- v0 y' O; i# s Breaking the problem into smaller instances of the same problem. 4 n, {* f" b9 U$ [! u \% T6 ~( v. [+ S* l8 ~% w
Solving the smallest instance directly (base case). , P9 p4 K" I, H5 D8 y8 t. R & i6 L: t: I/ H6 g# c2 r! L Combining the results of smaller instances to solve the larger problem." {% d, G8 A: ^* y9 S
4 e$ C! [+ \- N1 J2 P0 ZComponents of a Recursive Function, N0 m4 t, Z- @5 m p
9 S5 }4 j/ A' j( q% \2 p Base Case: ; P2 w2 V% M) E2 Z& C# S8 Q- G, H
This is the simplest, smallest instance of the problem that can be solved directly without further recursion. 3 ?6 k( s+ ]8 x, D6 z 3 l* e* B2 }& Z# E3 O, @2 U It acts as the stopping condition to prevent infinite recursion. 5 `. r% x' I8 q) e ; t; x) s- Z" X' B4 {9 [2 j. ^ Example: In calculating the factorial of a number, the base case is factorial(0) = 1. * I' N5 Q/ I$ k0 \! ^ & g0 i3 Y, U1 o3 M5 |3 ?# ~+ s3 w) ^3 t Recursive Case:7 @9 A/ N. ]" X' Y3 s
3 A% s, e- G3 g) A' X9 m This is where the function calls itself with a smaller or simpler version of the problem.4 w& ?9 v$ x! W( v% g1 d
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Example: For factorial, the recursive case is factorial(n) = n * factorial(n-1). + Y8 w. E9 G/ X5 f1 S) V & E* b' g! l+ u/ |Example: Factorial Calculation 2 ~- o* j5 z/ |3 Y9 y4 g * A" o- F* r, z8 k0 ZThe 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:7 Z3 i- c3 {+ L! E4 y. @ b
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Base case: 0! = 1 7 R1 q6 K/ w( P) H / U v ^* w# P. {3 c Recursive case: n! = n * (n-1)! L# Q1 F& e" _+ e
* A6 q! L# P% K/ V1 XHere’s how it looks in code (Python): + X I/ Y5 b$ Z, T" a7 upython , P) s$ u2 C( R( k/ g. _8 E7 W1 ? + z5 A9 }; H# y' C, ]% v4 r: t6 X8 @1 x2 Z) k r4 @% @
def factorial(n):+ d) a, {3 F1 p0 h+ W% S0 E9 W( `1 p
# Base case 0 D: x. j$ f4 X4 H1 _% ^) @- Y8 ^ if n == 0: ; l3 i# s( }$ |$ @* ~" A return 1, W2 o6 m) W8 P+ A u" h
# Recursive case , q' ]& m8 X' v+ _" K0 U else: * r! @( g* r% L5 t& ?: O return n * factorial(n - 1), I3 i, z0 w& G& ^; P( P
+ Q% q) L. \" _9 L+ Y: @+ S# Example usage% K( w8 H/ k6 N
print(factorial(5)) # Output: 120 " k% ^) k1 Y- [6 {& q( p- k. X) F+ R8 S% i' ~) D
How Recursion Works 4 w4 C u+ X6 J8 P) K% @( [3 d1 d) @$ J. V+ M. o
The function keeps calling itself with smaller inputs until it reaches the base case. & f: {7 N! Q% h0 z1 S( u. e 8 c- Y% D k g" V1 t1 { Once the base case is reached, the function starts returning values back up the call stack.6 Z/ B' ^( ?; H
/ l$ c4 N1 U$ L' j' W0 v. ~. |8 @Advantages of Recursion# S, m0 P, s r$ T4 L
0 u1 v8 g5 M a: Z1 I5 l 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). ; X# A% w, f/ D ; ]) i* ]# m. V8 T; B Readability: Recursive code can be more readable and concise compared to iterative solutions.4 I9 X* u, r9 D' _0 Y
/ p/ j7 L' s) MDisadvantages of Recursion0 f1 a$ c1 p/ D- B# I, a4 `
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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. - r* o7 B% C* E& x6 r1 Y( J V9 x- o7 [/ S; g, c& d
Inefficiency: Some problems can be solved more efficiently using iteration (e.g., Fibonacci sequence without memoization).5 B: Y+ d$ O8 z& F. ?5 Z0 ]7 p
2 e1 x# E' N5 q' E% s5 y/ P+ |- s' JWhen to Use Recursion ; s* P5 a6 A) Z. c6 f6 q3 [/ V1 z% ^' Q: ~" W
Problems that can be broken down into smaller, similar subproblems (e.g., tree traversals, sorting algorithms like quicksort and mergesort). 7 s. a" Q2 N% N( E7 C1 x) c; L . q2 x6 S: @, f! l Problems with a clear base case and recursive case. ! D) j" S" E n4 P/ A" p/ j' l: j9 d9 `
Example: Fibonacci Sequence & w" @9 k9 E1 Y1 y, q $ k7 D* G" @ g) [8 i5 t) \The Fibonacci sequence is another classic example of recursion. Each number is the sum of the two preceding ones:- P1 L5 I1 d$ ^5 |2 r
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Base case: fib(0) = 0, fib(1) = 1! U- h0 ^8 O( N7 {
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Recursive case: fib(n) = fib(n-1) + fib(n-2) ! Q% ^2 q8 l# [! n, u8 x5 P8 N. H& l* s( _8 k$ U* j/ \
python * E8 t k% c! o4 b. j2 S& m# A, f% N. R% g, N
2 t& }3 V7 ?' O& \% [def fibonacci(n): " n4 k/ ^5 C3 W7 T # Base cases# ? ]% ?0 g* U1 R( s
if n == 0: $ O1 S7 b: N9 z9 K" _0 m! y return 0/ b6 ?6 r, j- R N1 x
elif n == 1: ! Y% ?" B1 j: _% o3 v2 O4 c return 18 C) d, x( A0 w1 u8 m
# Recursive case . @' z$ z( d+ L8 ?' t$ ^ else:; d4 o7 [( _- G% T
return fibonacci(n - 1) + fibonacci(n - 2) 1 L# w1 m" E. g& o \% Z' B- C, Q- `. p% b5 }
# Example usage ' \9 k6 L6 P0 O' l3 {print(fibonacci(6)) # Output: 8 a4 K6 B- l/ _: X' E
0 S9 l8 x$ t0 E v0 i& l! VTail Recursion : \! I$ y! Q8 f5 [" b$ D, ?: X" I2 Y
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). " e$ Q, O: z# x$ S% l+ s, j( D& Q5 L. a( a4 z3 l/ [
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。