摘要:我们经常会用到递归函数,但是如果递归深度太大时,往往导致栈溢出。而递归深度往往不太容易把握,所以比较安全一点的做法就是:用循环代替递归。文章最后的原文里面讲了如何用10步实现这个过程,相当精彩。本文翻译了这篇文章,并加了自己的一点注释和理解。
目录
- 简介
- 模拟函数的目的
- 递归和模拟函数的优缺点
- 用栈和循环代替递归的10个步骤
- 替代过程的几个简单例子
- 更多的例子
- 结论
- 参考
- 协议
1 简介
一般我们在进行排序(比如归并排序)或者树操作时会用到递归函数。但是如果递归深度达到一定程度以后,就会出现意想不到的结果比如堆栈溢出。虽然有很多有经验的开发者都知道了如何用循环函数或者栈加while循环来代替递归函数,以防止栈溢出,但我还是想分享一下这些方法,这或许会对初学者有很大的帮助。
2 模拟函数的目的
如果你正在使用递归函数,并且没有控制递归调用,而栈资源又比较有限,调用层次过深的时候就可能导致栈溢出/堆冲突。模拟函数的目的就是在堆中开辟区域来模拟栈的行为,这样你就能控制内存分配和流处理,从而避免栈溢出。如果能用循环函数来代替效果会更好,这是一个比较需要时间和经验来处理的事情,出于这些原因,这篇文章为初学者提供了一个简单的参考,怎样使用循环函数来替代递归函数,以防止栈溢出?
3 递归函数和模拟函数的优缺点
递归函数:
优点:算法比较直观。可以参考文章后面的例子
缺点:可能导致栈溢出或者堆冲突
你可以试试执行下面两个函数(后面的一个例子),IsEvenNumber(递归实现)和IsEvenNumber(模拟实现),他们在头文件"MutualRecursion.h"中声明。你可以将传入参数设定为10000,像下面这样:
#include "MutualRecursion.h" bool result = IsEvenNumberLoop(10000); // 成功返回 bool result2 = IsEvenNumber(10000); // 会发生堆栈溢出
有些人可能会问:如果我增加栈的容量不就可以避免栈溢出吗?好吧,这只是暂时的解决问题的办法,如果调用层次越来越深,很有可能会再次发生溢出。
模拟函数:
优点:能避免栈溢出或者堆冲突错误,能对过程和内存进行更好的控制
缺点:算法不是太直观,代码难以维护
4 用栈和循环代替递归的10个步骤
第一步
1 定义一个新的结构体Snapshot,用于保存递归结构中的一些数据和状态信息
2 在Snapshot内部需要包含的变量有以下几种:
A 一般当递归函数调用自身时,函数参数会发生变化。所以你需要包含变化的参数,引用除外。比如下面的例子中,参数n应该包含在结构体中,而retVal不需要。
void SomeFunc(int n, int &retVal);
B 阶段性变量"stage"(通常是一个用来转换到另一个处理分支的整形变量),详见第六条规则
C 函数调用返回以后还需要继续使用的局部变量(一般在二分递归和嵌套递归中很常见)
代码:
1 // Recursive Function "First rule" example 2 int SomeFunc(int n, int &retIdx) 3 { 4 ... 5 if(n>0) 6 { 7 int test = SomeFunc(n-1, retIdx); 8 test--; 9 ... 10 return test; 11 } 12 ... 13 return 0; 14 } 15 16 17 // Conversion to Iterative Function 18 int SomeFuncLoop(int n, int &retIdx) 19 { 20 // (First rule) 21 struct SnapShotStruct { 22 int n; // - parameter input 23 int test; // - local variable that will be used 24 // after returning from the function call 25 // - retIdx can be ignored since it is a reference. 26 int stage; // - Since there is process needed to be done 27 // after recursive call. (Sixth rule) 28 }; 29 ... 30 }
第二步
1 在函数的开头创建一个局部变量,这个值扮演了递归函数的返回函数角色。它相当于为每次递归调用保存一个临时值,因为C++函数只能有一种返回类型,如果递归函数的返回类型是void,你可以忽略这个局部变量。如果有缺省的返回值,就应该用缺省值初始化这个局部变量。
1 // Recursive Function "Second rule" example 2 int SomeFunc(int n, int &retIdx) 3 { 4 ... 5 if(n>0) 6 { 7 int test = SomeFunc(n-1, retIdx); 8 test--; 9 ... 10 return test; 11 } 12 ... 13 return 0; 14 } 15 16 // Conversion to Iterative Function 17 int SomeFuncLoop(int n, int &retIdx) 18 { 19 // (First rule) 20 struct SnapShotStruct { 21 int n; // - parameter input 22 int test; // - local variable that will be used 23 // after returning from the function call 24 // - retIdx can be ignored since it is a reference. 25 int stage; // - Since there is process needed to be done 26 // after recursive call. (Sixth rule) 27 }; 28 29 // (Second rule) 30 int retVal = 0; // initialize with default returning value 31 32 ... 33 // (Second rule) 34 return retVal; 35 }
第三步
创建一个栈用于保存“Snapshot”结构体类型变量
1 // Recursive Function "Third rule" example 2 3 // Conversion to Iterative Function 4 int SomeFuncLoop(int n, int &retIdx) 5 { 6 // (First rule) 7 struct SnapShotStruct { 8 int n; // - parameter input 9 int test; // - local variable that will be used 10 // after returning from the function call 11 // - retIdx can be ignored since it is a reference. 12 int stage; // - Since there is process needed to be done 13 // after recursive call. (Sixth rule) 14 }; 15 16 // (Second rule) 17 int retVal = 0; // initialize with default returning value 18 19 // (Third rule) 20 stack<SnapShotStruct> snapshotStack; 21 ... 22 // (Second rule) 23 return retVal; 24 }
第四步
创建一个新的”Snapshot”实例,然后将其中的参数等初始化,并将“Snapshot”实例压入栈
1 // Recursive Function "Fourth rule" example 2 3 // Conversion to Iterative Function 4 int SomeFuncLoop(int n, int &retIdx) 5 { 6 // (First rule) 7 struct SnapShotStruct { 8 int n; // - parameter input 9 int test; // - local variable that will be used 10 // after returning from the function call 11 // - retIdx can be ignored since it is a reference. 12 int stage; // - Since there is process needed to be done 13 // after recursive call. (Sixth rule) 14 }; 15 16 // (Second rule) 17 int retVal = 0; // initialize with default returning value 18 19 // (Third rule) 20 stack<SnapShotStruct> snapshotStack; 21 22 // (Fourth rule) 23 SnapShotStruct currentSnapshot; 24 currentSnapshot.n= n; // set the value as parameter value 25 currentSnapshot.test=0; // set the value as default value 26 currentSnapshot.stage=0; // set the value as initial stage 27 28 snapshotStack.push(currentSnapshot); 29 30 ... 31 // (Second rule) 32 return retVal; 33 }
第五步
写一个while循环,使其不断执行直到栈为空。在while循环的每一次迭代过程中,弹出”Snapshot“对象。
1 // Recursive Function "Fifth rule" example 2 3 // Conversion to Iterative Function 4 int SomeFuncLoop(int n, int &retIdx) 5 { 6 // (First rule) 7 struct SnapShotStruct { 8 int n; // - parameter input 9 int test; // - local variable that will be used 10 // after returning from the function call 11 // - retIdx can be ignored since it is a reference. 12 int stage; // - Since there is process needed to be done 13 // after recursive call. (Sixth rule) 14 }; 15 // (Second rule) 16 int retVal = 0; // initialize with default returning value 17 // (Third rule) 18 stack<SnapShotStruct> snapshotStack; 19 // (Fourth rule) 20 SnapShotStruct currentSnapshot; 21 currentSnapshot.n= n; // set the value as parameter value 22 currentSnapshot.test=0; // set the value as default value 23 currentSnapshot.stage=0; // set the value as initial stage 24 snapshotStack.push(currentSnapshot); 25 // (Fifth rule) 26 while(!snapshotStack.empty()) 27 { 28 currentSnapshot=snapshotStack.top(); 29 snapshotStack.pop(); 30 ... 31 } 32 // (Second rule) 33 return retVal; 34 }
第六步
- 将当前阶段一分为二(针对当前只有单一递归调用的情形)。第一个阶段代表了下一次递归调用之前的情况,第二阶段代表了下一次递归调用完成并返回之后的情况(返回值已经被保存,并在此之前被累加)。
- 如果当前阶段有两次递归调用,就必须分为3个阶段。阶段1:第一次调用返回之前,阶段2:阶段1执行的调用过程。阶段3:第二次调用返回之前。
- 如果当前阶段有三次递归调用,就必须至少分为4个阶段。
- 依次类推。
1 // Recursive Function "Sixth rule" example 2 int SomeFunc(int n, int &retIdx) 3 { 4 ... 5 if(n>0) 6 { 7 int test = SomeFunc(n-1, retIdx); 8 test--; 9 ... 10 return test; 11 } 12 ... 13 return 0; 14 } 15 16 // Conversion to Iterative Function 17 int SomeFuncLoop(int n, int &retIdx) 18 { 19 // (First rule) 20 struct SnapShotStruct { 21 int n; // - parameter input 22 int test; // - local variable that will be used 23 // after returning from the function call 24 // - retIdx can be ignored since it is a reference. 25 int stage; // - Since there is process needed to be done 26 // after recursive call. (Sixth rule) 27 }; 28 // (Second rule) 29 int retVal = 0; // initialize with default returning value 30 // (Third rule) 31 stack<SnapShotStruct> snapshotStack; 32 // (Fourth rule) 33 SnapShotStruct currentSnapshot; 34 currentSnapshot.n= n; // set the value as parameter value 35 currentSnapshot.test=0; // set the value as default value 36 currentSnapshot.stage=0; // set the value as initial stage 37 snapshotStack.push(currentSnapshot); 38 // (Fifth rule) 39 while(!snapshotStack.empty()) 40 { 41 currentSnapshot=snapshotStack.top(); 42 snapshotStack.pop(); 43 // (Sixth rule) 44 switch( currentSnapshot.stage) 45 { 46 case 0: 47 ... // before ( SomeFunc(n-1, retIdx); ) 48 break; 49 case 1: 50 ... // after ( SomeFunc(n-1, retIdx); ) 51 break; 52 } 53 } 54 // (Second rule) 55 return retVal; 56 }
第七步
根据阶段变量stage的值切换到相应的处理流程并处理相关过程。
1 // Recursive Function "Seventh rule" example 2 int SomeFunc(int n, int &retIdx) 3 { 4 ... 5 if(n>0) 6 { 7 int test = SomeFunc(n-1, retIdx); 8 test--; 9 ... 10 return test; 11 } 12 ... 13 return 0; 14 } 15 16 // Conversion to Iterative Function 17 int SomeFuncLoop(int n, int &retIdx) 18 { 19 // (First rule) 20 struct SnapShotStruct { 21 int n; // - parameter input 22 int test; // - local variable that will be used 23 // after returning from the function call 24 // - retIdx can be ignored since it is a reference. 25 int stage; // - Since there is process needed to be done 26 // after recursive call. (Sixth rule) 27 }; 28 29 // (Second rule) 30 int retVal = 0; // initialize with default returning value 31 32 // (Third rule) 33 stack<SnapShotStruct> snapshotStack; 34 35 // (Fourth rule) 36 SnapShotStruct currentSnapshot; 37 currentSnapshot.n= n; // set the value as parameter value 38 currentSnapshot.test=0; // set the value as default value 39 currentSnapshot.stage=0; // set the value as initial stage 40 41 snapshotStack.push(currentSnapshot); 42 43 // (Fifth rule) 44 while(!snapshotStack.empty()) 45 { 46 currentSnapshot=snapshotStack.top(); 47 snapshotStack.pop(); 48 49 // (Sixth rule) 50 switch( currentSnapshot.stage) 51 { 52 case 0: 53 // (Seventh rule) 54 if( currentSnapshot.n>0 ) 55 { 56 ... 57 } 58 ... 59 break; 60 case 1: 61 // (Seventh rule) 62 currentSnapshot.test = retVal; 63 currentSnapshot.test--; 64 ... 65 break; 66 } 67 } 68 // (Second rule) 69 return retVal; 70 }
第八步
如果递归有返回值,将这个值保存下来放在临时变量里面,比如retVal。当循环结束时,这个临时变量的值就是整个递归处理的结果。
1 // Recursive Function "Eighth rule" example 2 int SomeFunc(int n, int &retIdx) 3 { 4 ... 5 if(n>0) 6 { 7 int test = SomeFunc(n-1, retIdx); 8 test--; 9 ... 10 return test; 11 } 12 ... 13 return 0; 14 } 15 16 // Conversion to Iterative Function 17 int SomeFuncLoop(int n, int &retIdx) 18 { 19 // (First rule) 20 struct SnapShotStruct { 21 int n; // - parameter input 22 int test; // - local variable that will be used 23 // after returning from the function call 24 // - retIdx can be ignored since it is a reference. 25 int stage; // - Since there is process needed to be done 26 // after recursive call. (Sixth rule) 27 }; 28 // (Second rule) 29 int retVal = 0; // initialize with default returning value 30 // (Third rule) 31 stack<SnapShotStruct> snapshotStack; 32 // (Fourth rule) 33 SnapShotStruct currentSnapshot; 34 currentSnapshot.n= n; // set the value as parameter value 35 currentSnapshot.test=0; // set the value as default value 36 currentSnapshot.stage=0; // set the value as initial stage 37 snapshotStack.push(currentSnapshot); 38 // (Fifth rule) 39 while(!snapshotStack.empty()) 40 { 41 currentSnapshot=snapshotStack.top(); 42 snapshotStack.pop(); 43 // (Sixth rule) 44 switch( currentSnapshot.stage) 45 { 46 case 0: 47 // (Seventh rule) 48 if( currentSnapshot.n>0 ) 49 { 50 ... 51 } 52 ... 53 // (Eighth rule) 54 retVal = 0 ; 55 ... 56 break; 57 case 1: 58 // (Seventh rule) 59 currentSnapshot.test = retVal; 60 currentSnapshot.test--; 61 ... 62 // (Eighth rule) 63 retVal = currentSnapshot.test; 64 ... 65 break; 66 } 67 } 68 // (Second rule) 69 return retVal; 70 }
第九步
如果递归函数有“return”关键字,你应该在while循环里面用“continue”代替。如果return了一个返回值,你应该在循环里面保存下来(步骤8),然后return。大部分情况下,步骤九是可选的,但是它能帮助你避免逻辑错误。
1 // Recursive Function "Ninth rule" example 2 int SomeFunc(int n, int &retIdx) 3 { 4 ... 5 if(n>0) 6 { 7 int test = SomeFunc(n-1, retIdx); 8 test--; 9 ... 10 return test; 11 } 12 ... 13 return 0; 14 } 15 16 // Conversion to Iterative Function 17 int SomeFuncLoop(int n, int &retIdx) 18 { 19 // (First rule) 20 struct SnapShotStruct { 21 int n; // - parameter input 22 int test; // - local variable that will be used 23 // after returning from the function call 24 // - retIdx can be ignored since it is a reference. 25 int stage; // - Since there is process needed to be done 26 // after recursive call. (Sixth rule) 27 }; 28 // (Second rule) 29 int retVal = 0; // initialize with default returning value 30 // (Third rule) 31 stack<SnapShotStruct> snapshotStack; 32 // (Fourth rule) 33 SnapShotStruct currentSnapshot; 34 currentSnapshot.n= n; // set the value as parameter value 35 currentSnapshot.test=0; // set the value as default value 36 currentSnapshot.stage=0; // set the value as initial stage 37 snapshotStack.push(currentSnapshot); 38 // (Fifth rule) 39 while(!snapshotStack.empty()) 40 { 41 currentSnapshot=snapshotStack.top(); 42 snapshotStack.pop(); 43 // (Sixth rule) 44 switch( currentSnapshot.stage) 45 { 46 case 0: 47 // (Seventh rule) 48 if( currentSnapshot.n>0 ) 49 { 50 ... 51 } 52 ... 53 // (Eighth rule) 54 retVal = 0 ; 55 56 // (Ninth rule) 57 continue; 58 break; 59 case 1: 60 // (Seventh rule) 61 currentSnapshot.test = retVal; 62 currentSnapshot.test--; 63 ... 64 // (Eighth rule) 65 retVal = currentSnapshot.test; 66 67 // (Ninth rule) 68 continue; 69 break; 70 } 71 } 72 // (Second rule) 73 return retVal; 74 }
第十步
为了模拟下一次递归函数的调用,你必须在当前循环函数里面再生成一个新的“Snapshot”结构体作为下一次调用的快照,初始化其参数以后压入栈,并“continue”。如果当前调用在执行完成后还有一些事情需要处理,那么更改它的阶段状态“stage”到相应的过程,并在new Snapshot压入之前,把本次的“Snapshot”压入。
1 // Recursive Function "Tenth rule" example 2 int SomeFunc(int n, int &retIdx) 3 { 4 ... 5 if(n>0) 6 { 7 int test = SomeFunc(n-1, retIdx); 8 test--; 9 ... 10 return test; 11 } 12 ... 13 return 0; 14 } 15 16 // Conversion to Iterative Function 17 int SomeFuncLoop(int n, int &retIdx) 18 { 19 // (First rule) 20 struct SnapShotStruct { 21 int n; // - parameter input 22 int test; // - local variable that will be used 23 // after returning from the function call 24 // - retIdx can be ignored since it is a reference. 25 int stage; // - Since there is process needed to be done 26 // after recursive call. (Sixth rule) 27 }; 28 // (Second rule) 29 int retVal = 0; // initialize with default returning value 30 // (Third rule) 31 stack<SnapShotStruct> snapshotStack; 32 // (Fourth rule) 33 SnapShotStruct currentSnapshot; 34 currentSnapshot.n= n; // set the value as parameter value 35 currentSnapshot.test=0; // set the value as default value 36 currentSnapshot.stage=0; // set the value as initial stage 37 snapshotStack.push(currentSnapshot); 38 // (Fifth rule) 39 while(!snapshotStack.empty()) 40 { 41 currentSnapshot=snapshotStack.top(); 42 snapshotStack.pop(); 43 // (Sixth rule) 44 switch( currentSnapshot.stage) 45 { 46 case 0: 47 // (Seventh rule) 48 if( currentSnapshot.n>0 ) 49 { 50 // (Tenth rule) 51 currentSnapshot.stage = 1; // - current snapshot need to process after 52 // returning from the recursive call 53 snapshotStack.push(currentSnapshot); // - this MUST pushed into stack before 54 // new snapshot! 55 // Create a new snapshot for calling itself 56 SnapShotStruct newSnapshot; 57 newSnapshot.n= currentSnapshot.n-1; // - give parameter as parameter given 58 // when calling itself 59 // ( SomeFunc(n-1, retIdx) ) 60 newSnapshot.test=0; // - set the value as initial value 61 newSnapshot.stage=0; // - since it will start from the 62 // beginning of the function, 63 // give the initial stage 64 snapshotStack.push(newSnapshot); 65 continue; 66 } 67 ... 68 // (Eighth rule) 69 retVal = 0 ; 70 71 // (Ninth rule) 72 continue; 73 break; 74 case 1: 75 // (Seventh rule) 76 currentSnapshot.test = retVal; 77 currentSnapshot.test--; 78 ... 79 // (Eighth rule) 80 retVal = currentSnapshot.test; 81 // (Ninth rule) 82 continue; 83 break; 84 } 85 } 86 // (Second rule) 87 return retVal; 88 }
5 替代过程的几个简单例子
以下几个例子均在vs2008环境下开发,主要包含了:
(1)线性递归
1 #ifndef __LINEAR_RECURSION_H__ 2 #define __LINEAR_RECURSION_H__ 3 4 #include <stack> 5 using namespace std; 6 7 /** 8 * rief 求n的阶乘 9 * para 10 * eturn 11 * ote result = n! 递归实现 12 */ 13 int Fact(long n) 14 { 15 if(0>n) 16 return -1; 17 if(0 == n) 18 return 1; 19 else 20 { 21 return ( n* Fact(n-1)); 22 } 23 } 24 25 /** 26 * rief 求n的阶乘 27 * para 28 * eturn 29 * ote result = n! 循环实现 30 */ 31 int FactLoop(long n) 32 { 33 // (步骤1) 34 struct SnapShotStruct // 快照结构体局部声明 35 { 36 long inputN; // 会改变的参数 37 // 没有局部变量 38 int stage; // 阶段变量用于快照跟踪 39 } ; 40 41 // (步骤2) 42 int returnVal; // 用于保存当前调用返回值 43 44 // (步骤3) 45 stack<SnapShotStruct> snapshotStack; 46 47 // (步骤4) 48 SnapShotStruct currentSnapshot; 49 currentSnapshot.inputN=n; 50 currentSnapshot.stage=0; // 阶段变量初始化 51 52 snapshotStack.push(currentSnapshot); 53 54 // (步骤5) 55 while(!snapshotStack.empty()) 56 { 57 currentSnapshot=snapshotStack.top(); 58 snapshotStack.pop(); 59 60 // (步骤6) 61 switch(currentSnapshot.stage) 62 { 63 // (步骤7) 64 case 0: 65 if(0>currentSnapshot.inputN) 66 { 67 // (步骤8 & 步骤9) 68 returnVal = -1; 69 continue; 70 } 71 if(0 == currentSnapshot.inputN) 72 { 73 // (步骤8 & 步骤9) 74 returnVal = 1; 75 continue; 76 } 77 else 78 { 79 // (步骤10) 80 81 // 返回 ( n* Fact(n-1)); 分为2步: 82 // (第一步调用自身,第二步用返回值乘以当前n值) 83 // 这里我们拍下快照. 84 currentSnapshot.stage=1; // 当前的快照表示正在被处理,并等待自身调用结果返回,所以赋值为1 85 86 snapshotStack.push(currentSnapshot); 87 88 // 创建一个新的快照,用于调用自身 89 SnapShotStruct newSnapshot; 90 newSnapshot.inputN= currentSnapshot.inputN -1 ; // 初始化参数 91 92 newSnapshot.stage = 0 ; // 从头开始执行自身,所以赋值stage==0 93 94 snapshotStack.push(newSnapshot); 95 continue; 96 97 } 98 break; 99 // (步骤7) 100 case 1: 101 102 // (步骤8) 103 104 returnVal = currentSnapshot.inputN * returnVal; 105 106 // (步骤9) 107 continue; 108 break; 109 } 110 } 111 112 // (步骤2) 113 return returnVal; 114 } 115 #endif //__LINEAR_RECURSION_H__
(2)二分递归
1 #ifndef __BINARY_RECURSION_H__ 2 #define __BINARY_RECURSION_H__ 3 4 #include <stack> 5 using namespace std; 6 7 /** 8 * function FibNum 9 * rief 求斐波纳契数列 10 * para 11 * eturn 12 * ote 递归实现 13 */ 14 int FibNum(int n) 15 { 16 if (n < 1) 17 return -1; 18 if (1 == n || 2 == n) 19 return 1; 20 21 // 这里可以看成是 22 //int addVal = FibNum( n - 1); 23 // addVal += FibNum(n - 2); 24 // return addVal; 25 return FibNum(n - 1) + FibNum(n - 2); 26 } 27 /** 28 * function FibNumLoop 29 * rief 求斐波纳契数列 30 * para 31 * eturn 32 * ote 循环实现 33 */ 34 int FibNumLoop(int n) 35 { 36 // (步骤1) 37 struct SnapShotStruct // 快照结构体局部声明 38 { 39 int inputN; // 会改变的参数 40 int addVal; // 局部变量 41 int stage; // 阶段变量用于快照跟踪 42 43 }; 44 45 // (步骤2) 46 int returnVal; // 用于保存当前调用返回值 47 48 // (步骤3) 49 stack<SnapShotStruct> snapshotStack; 50 51 // (步骤4) 52 SnapShotStruct currentSnapshot; 53 currentSnapshot.inputN=n; 54 currentSnapshot.stage=0; // 阶段变量初始化 55 56 snapshotStack.push(currentSnapshot); 57 58 // (步骤5) 59 while(!snapshotStack.empty()) 60 { 61 currentSnapshot=snapshotStack.top(); 62 snapshotStack.pop(); 63 64 // (步骤6) 65 switch(currentSnapshot.stage) 66 { 67 // (步骤7) 68 case 0: 69 if(currentSnapshot.inputN<1) 70 { 71 // (步骤8 & 步骤9) 72 returnVal = -1; 73 continue; 74 } 75 if(currentSnapshot.inputN == 1 || currentSnapshot.inputN == 2 ) 76 { 77 // (步骤8 & 步骤9) 78 returnVal = 1; 79 continue; 80 } 81 else 82 { 83 // (步骤10) 84 85 // 返回 ( FibNum(n - 1) + FibNum(n - 2)); 相当于两步 86 // (第一次调用参数是 n-1, 第二次调用参数 n-2) 87 // 这里我们拍下快照,分成2个阶段 88 currentSnapshot.stage=1; // 当前的快照表示正在被处理,并等待自身调用结果返回,所以赋值为1 89 90 snapshotStack.push(currentSnapshot); 91 92 // 创建一个新的快照,用于调用自身 93 SnapShotStruct newSnapshot; 94 newSnapshot.inputN= currentSnapshot.inputN -1 ; //初始化参数 FibNum(n - 1) 95 96 newSnapshot.stage = 0 ; 97 snapshotStack.push(newSnapshot); 98 continue; 99 100 } 101 break; 102 // (步骤7) 103 case 1: 104 105 // (步骤10) 106 107 currentSnapshot.addVal = returnVal; 108 currentSnapshot.stage=2; // 当前的快照正在被处理,并等待的自身调用结果,所以阶段变量变成2 109 110 snapshotStack.push(currentSnapshot); 111 112 // 创建一个新的快照,用于调用自身 113 SnapShotStruct newSnapshot; 114 newSnapshot.inputN= currentSnapshot.inputN - 2 ; // 初始化参数 FibNum(n - 2) 115 newSnapshot.stage = 0 ; // 从头开始执行,阶段变量赋值为0 116 117 snapshotStack.push(newSnapshot); 118 continue; 119 break; 120 case 2: 121 // (步骤8) 122 returnVal = currentSnapshot.addVal + returnVal; // actual addition of ( FibNum(n - 1) + FibNum(n - 2) ) 123 124 // (步骤9) 125 continue; 126 break; 127 } 128 } 129 130 // (步骤2) 131 return returnVal; 132 } 133 134 135 #endif //__BINARY_RECURSION_H__
(3)尾递归
1 #ifndef __TAIL_RECURSION_H__ 2 #define __TAIL_RECURSION_H__ 3 4 #include <stack> 5 using namespace std; 6 7 /** 8 * function FibNum2 9 * rief 2阶裴波那契序列 10 * para 11 * eturn 12 * ote 递归实现 f0 = x, f1 = y, fn=fn-1+fn-2, n=k,k+1,... 13 */ 14 int FibNum2(int n, int x, int y) 15 { 16 if (1 == n) 17 { 18 return y; 19 } 20 else 21 { 22 return FibNum2(n-1, y, x+y); 23 } 24 } 25 /** 26 * function FibNum2Loop 27 * rief 2阶裴波那契序列 28 * para 29 * eturn 30 * ote 循环实现 在尾递归中, 递归调用后除了返回没有任何其它的操作, 所以在变为循环时,不需要stage变量 31 */ 32 int FibNum2Loop(int n, int x, int y) 33 { 34 // (步骤1) 35 struct SnapShotStruct 36 { 37 int inputN; // 会改变的参数 38 int inputX; // 会改变的参数 39 int inputY; // 会改变的参数 40 // 没有局部变量 41 }; 42 43 // (步骤2) 44 int returnVal; 45 46 // (步骤3) 47 stack<SnapShotStruct> snapshotStack; 48 49 // (步骤4) 50 SnapShotStruct currentSnapshot; 51 currentSnapshot.inputN = n; 52 currentSnapshot.inputX = x; 53 currentSnapshot.inputY = y; 54 55 snapshotStack.push(currentSnapshot); 56 57 // (步骤5) 58 while(!snapshotStack.empty()) 59 { 60 currentSnapshot=snapshotStack.top(); 61 snapshotStack.pop(); 62 63 if(currentSnapshot.inputN == 1) 64 { 65 // (步骤8 & 步骤9) 66 returnVal = currentSnapshot.inputY; 67 continue; 68 } 69 else 70 { 71 // (步骤10) 72 73 // 创建新快照 74 SnapShotStruct newSnapshot; 75 newSnapshot.inputN= currentSnapshot.inputN -1 ; // 初始化,调用( FibNum(n-1, y, x+y) ) 76 newSnapshot.inputX= currentSnapshot.inputY; 77 newSnapshot.inputY= currentSnapshot.inputX + currentSnapshot.inputY; 78 snapshotStack.push(newSnapshot); 79 continue; 80 } 81 } 82 // (步骤2) 83 return returnVal; 84 } 85 86 #endif //__TAIL_RECURSION_H__
(4)互递归
1 #ifndef __MUTUAL_RECURSION_H__ 2 #define __MUTUAL_RECURSION_H__ 3 #include <stack> 4 using namespace std; 5 6 bool IsEvenNumber(int n);//判断是否是偶数 7 bool IsOddNumber(int n);//判断是否是奇数 8 bool isOddOrEven(int n, int stage);//判断是否是奇数或偶数 9 10 /****************************************************/ 11 //互相调用的递归实现 12 bool IsOddNumber(int n) 13 { 14 // 终止条件 15 if (0 == n) 16 return false; 17 else 18 // 互相调用函数的递归调用 19 return IsEvenNumber(n - 1); 20 } 21 22 bool IsEvenNumber(int n) 23 { 24 // 终止条件 25 if (0 == n) 26 return true; 27 else 28 // 互相调用函数的递归调用 29 return IsOddNumber(n - 1); 30 } 31 32 33 /*************************************************/ 34 //互相调用的循环实现 35 bool IsOddNumberLoop(int n) 36 { 37 return isOddOrEven(n , 0); 38 } 39 40 bool IsEvenNumberLoop(int n) 41 { 42 return isOddOrEven(n , 1); 43 } 44 45 bool isOddOrEven(int n, int stage) 46 { 47 // (步骤1) 48 struct SnapShotStruct 49 { 50 int inputN; // 会改变的参数 51 int stage; 52 // 没有局部变量 53 }; 54 55 // (步骤2) 56 bool returnVal; 57 58 // (步骤3) 59 stack<SnapShotStruct> snapshotStack; 60 61 // (步骤4) 62 SnapShotStruct currentSnapshot; 63 currentSnapshot.inputN = n; 64 currentSnapshot.stage = stage; 65 66 snapshotStack.push(currentSnapshot); 67 68 // (步骤5) 69 while(!snapshotStack.empty()) 70 { 71 currentSnapshot=snapshotStack.top(); 72 snapshotStack.pop(); 73 74 // (步骤6) 75 switch(currentSnapshot.stage) 76 { 77 // (步骤7) 78 // bool IsOddNumber(int n) 79 case 0: 80 // 终止条件 81 if (0 == currentSnapshot.inputN) 82 { 83 // (步骤8 & 步骤9) 84 returnVal = false; 85 continue; 86 } 87 else 88 { 89 // (步骤10) 90 91 // 模拟互调用的递归调用 92 93 // 创建新的快照 94 SnapShotStruct newSnapshot; 95 newSnapshot.inputN= currentSnapshot.inputN - 1; // 初始化参数 96 // 调用 ( IsEvenNumber(n - 1) ) 97 newSnapshot.stage= 1; 98 snapshotStack.push(newSnapshot); 99 continue; 100 } 101 102 break; 103 // (步骤7) 104 // bool IsEvenNumber(int n) 105 case 1: 106 // 终止条件 107 if (0 == currentSnapshot.inputN) 108 { 109 // (步骤8 & 步骤9) 110 returnVal = true; 111 continue; 112 } 113 else 114 { 115 // (步骤10) 116 117 // 模拟互调用的递归调用 118 119 // 创建新的快照 120 SnapShotStruct newSnapshot; 121 newSnapshot.inputN= currentSnapshot.inputN - 1; // 122 // calling itself ( IsEvenNumber(n - 1) ) 123 newSnapshot.stage= 0; 124 snapshotStack.push(newSnapshot); 125 continue; 126 } 127 break; 128 } 129 130 } 131 // (步骤2) 132 return returnVal; 133 } 134 135 #endif //__MUTUAL_RECURSION_H__
(5)嵌套递归
1 #ifndef __NESTED_RECURSION_H__ 2 #define __NESTED_RECURSION_H__ 3 #include <stack> 4 using namespace std; 5 6 int Ackermann(int x, int y) 7 { 8 // 终止条件 9 if (0 == x) 10 { 11 return y + 1; 12 } 13 // 错误处理条件 14 if (x < 0 || y < 0) 15 { 16 return -1; 17 } 18 // 线性方法的递归调用 19 else if (x > 0 && 0 == y) 20 { 21 return Ackermann(x-1, 1); 22 } 23 // 嵌套方法的递归调用 24 else 25 { 26 //可以看成是: 27 // int midVal = Ackermann(x, y-1); 28 // return Ackermann(x-1, midVal); 29 return Ackermann(x-1, Ackermann(x, y-1)); 30 } 31 } 32 33 34 35 int AckermannLoop(int x, int y) 36 { 37 // (步骤1) 38 struct SnapShotStruct 39 { 40 int inputX; // 会改变的参数 41 int inputY; // 会改变的参数 42 int stage; 43 // 没有局部变量 44 }; 45 46 // (步骤2) 47 int returnVal; 48 49 // (步骤3) 50 stack<SnapShotStruct> snapshotStack; 51 52 // (步骤4) 53 SnapShotStruct currentSnapshot; 54 currentSnapshot.inputX = x; 55 currentSnapshot.inputY = y; 56 currentSnapshot.stage = 0; 57 58 snapshotStack.push(currentSnapshot); 59 60 // (步骤5) 61 while(!snapshotStack.empty()) 62 { 63 currentSnapshot=snapshotStack.top(); 64 snapshotStack.pop(); 65 66 // (步骤6) 67 switch(currentSnapshot.stage) 68 { 69 // (步骤7) 70 case 0: 71 // 终止条件 72 if(currentSnapshot.inputX == 0) 73 { 74 // (步骤8 & 步骤9) 75 returnVal = currentSnapshot.inputY + 1; 76 continue; // 这里必须返回 77 } 78 // 错误处理条件 79 if (currentSnapshot.inputX < 0 || currentSnapshot.inputY < 0) 80 { 81 // (步骤8 & 步骤9) 82 returnVal = -1; 83 continue; // 这里必须返回 84 } 85 // 线性方法的递归调用 86 else if (currentSnapshot.inputX > 0 && 0 == currentSnapshot.inputY) 87 { 88 // (步骤10) 89 90 // 创建新快照 91 SnapShotStruct newSnapshot; 92 newSnapshot.inputX= currentSnapshot.inputX - 1; // 参数设定 calling itself ( Ackermann(x-1, 1) ) 93 newSnapshot.inputY= 1; // 参数设定 calling itself ( Ackermann(x-1, 1) ) 94 newSnapshot.stage= 0; 95 snapshotStack.push(newSnapshot); 96 continue; 97 } 98 // Recursive call by Nested method 99 else 100 { 101 // (步骤10) 102 103 currentSnapshot.stage=1; 104 snapshotStack.push(currentSnapshot); 105 106 // 创建新快照 107 SnapShotStruct newSnapshot; 108 newSnapshot.inputX= currentSnapshot.inputX; //参数设定calling itself ( Ackermann(x, y-1) ) 109 newSnapshot.inputY= currentSnapshot.inputY - 1; //参数设定calling itself ( Ackermann(x, y-1) ) 110 newSnapshot.stage = 0; 111 snapshotStack.push(newSnapshot); 112 continue; 113 } 114 break; 115 case 1: 116 // (步骤10) 117 118 // 创建新快照 119 SnapShotStruct newSnapshot; 120 newSnapshot.inputX= currentSnapshot.inputX - 1; // 设定参数calling itself ( Ackermann(x-1, Ackermann(x, y-1)) ) 121 newSnapshot.inputY= returnVal; // 设定参数calling itself ( Ackermann(x-1, Ackermann(x, y-1)) ) 122 newSnapshot.stage = 0; 123 snapshotStack.push(newSnapshot); 124 continue; 125 break; 126 } 127 } 128 // (步骤2) 129 return returnVal; 130 } 131 #endif //__NESTED_RECURSION_H__
测试代码:
1 #include <tchar.h> 2 #include "BinaryRecursion.h" 3 #include "LinearRecursion.h" 4 #include "MutualRecursion.h" 5 #include "NestedRecursion.h" 6 #include "TailRecursion.h" 7 8 9 int _tmain(int argc,_TCHAR argv[] ) 10 { 11 // Binary Recursion 12 int result = FibNum(10); 13 int result2 = FibNumLoop(10); 14 15 printf("FibNum(10) = %d ",result); 16 printf("FibNumLoop(10) = %d ",result2); 17 18 19 // Linear Recursion 20 result = Fact(10); 21 result2 = FactLoop(10); 22 23 printf("Fact(10) = %d ",result); 24 printf("FactLoop(10) = %d ",result2); 25 26 27 // Tail Recursion 28 result = FibNum2(10,5,4); 29 result2 = FibNum2Loop(10,5,4); 30 31 printf("FibNum2(10,5,4) = %d ",result); 32 printf("FibNumLoop2(10,5,4) = %d ",result2); 33 34 35 // Mutual Recursion 36 bool bResult = IsOddNumber(10); 37 bool bResult2 = IsOddNumberLoop(10); 38 39 bool bResult3 = IsEvenNumber(10); 40 bool bResult4 = IsEvenNumberLoop(10); 41 42 printf("IsOddNumber(10) = %d ",(int)bResult); 43 printf("IsOddNumberLoop(10) = %d ",(int)bResult2); 44 printf("IsEvenNumber(10) = %d ",(int)bResult3); 45 printf("IsEvenNumberLoop(10) = %d ",(int)bResult4); 46 47 48 // Nested Recursion 49 result = Ackermann(3,2); 50 result2 = AckermannLoop(3,2); 51 52 printf("Ackermann(3,2) = %d ",result); 53 printf("AckermannLoop(3,2) = %d ",result2); 54 55 while(1){} 56 return 0; 57 }
6 更多的例子
7 结论
我的结论就是在c/c++或者Java代码中,尽量避免用递归。但是正如你看到的,递归容易理解,但是容易导致栈溢出。虽然循环版本的函数不会增加代码可读性和提升性能,但是它能有效的避免冲突或未定义行为。正如我开头所说,我的做法通常是在代码中写两份代码,一份递归,一份循环的。前者用于理解代码,后者用于实际的运行和测试用。如果你对于自己代码中使用这两种代码的利弊很清楚,你可以选择你自己的方式。
8 参考
9 License
原文:http://www.codeproject.com/Articles/418776/How-to-replace-recursive-functions-using-stack-and
以上就是原文的一些内容,感谢原作者Woong Gyu La。
这篇文章中的代码我在调式过程中,发现了一个问题:循环版本的函数在执行效率方面存在问题。以后再改