泽州网站设计,深圳荷坳网站建设公司,如何把自己网站推广出去,商务网站要怎么做摘要#xff1a;我们经常会用到递归函数#xff0c;但是如果递归深度太大时#xff0c;往往导致栈溢出。而递归深度往往不太容易把握#xff0c;所以比较安全一点的做法就是#xff1a;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#xff0c;相当精彩…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 example2 int SomeFunc(int n, int retIdx)3 {4 ...5 if(n0)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 } View Code 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 example2 int SomeFunc(int n, int retIdx)3 {4 ...5 if(n0)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 } View Code 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 example2 3 // Conversion to Iterative Function4 int SomeFuncLoop(int n, int retIdx)5 {6 // (First rule)7 struct SnapShotStruct {8 int n; // - parameter input9 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 stackSnapShotStruct snapshotStack;
21 ...
22 // (Second rule)
23 return retVal;
24 } View Code 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 example2 3 // Conversion to Iterative Function4 int SomeFuncLoop(int n, int retIdx)5 {6 // (First rule)7 struct SnapShotStruct {8 int n; // - parameter input9 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 stackSnapShotStruct snapshotStack;
21
22 // (Fourth rule)
23 SnapShotStruct currentSnapshot;
24 currentSnapshot.n n; // set the value as parameter value
25 currentSnapshot.test0; // set the value as default value
26 currentSnapshot.stage0; // set the value as initial stage
27
28 snapshotStack.push(currentSnapshot);
29
30 ...
31 // (Second rule)
32 return retVal;
33 } View Code 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 example2 3 // Conversion to Iterative Function4 int SomeFuncLoop(int n, int retIdx)5 {6 // (First rule)7 struct SnapShotStruct {8 int n; // - parameter input9 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 stackSnapShotStruct snapshotStack;
19 // (Fourth rule)
20 SnapShotStruct currentSnapshot;
21 currentSnapshot.n n; // set the value as parameter value
22 currentSnapshot.test0; // set the value as default value
23 currentSnapshot.stage0; // set the value as initial stage
24 snapshotStack.push(currentSnapshot);
25 // (Fifth rule)
26 while(!snapshotStack.empty())
27 {
28 currentSnapshotsnapshotStack.top();
29 snapshotStack.pop();
30 ...
31 }
32 // (Second rule)
33 return retVal;
34 } View Code 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 example2 int SomeFunc(int n, int retIdx)3 {4 ...5 if(n0)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 stackSnapShotStruct snapshotStack;
32 // (Fourth rule)
33 SnapShotStruct currentSnapshot;
34 currentSnapshot.n n; // set the value as parameter value
35 currentSnapshot.test0; // set the value as default value
36 currentSnapshot.stage0; // set the value as initial stage
37 snapshotStack.push(currentSnapshot);
38 // (Fifth rule)
39 while(!snapshotStack.empty())
40 {
41 currentSnapshotsnapshotStack.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 } View Code 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 example2 int SomeFunc(int n, int retIdx)3 {4 ...5 if(n0)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 stackSnapShotStruct snapshotStack;
34
35 // (Fourth rule)
36 SnapShotStruct currentSnapshot;
37 currentSnapshot.n n; // set the value as parameter value
38 currentSnapshot.test0; // set the value as default value
39 currentSnapshot.stage0; // set the value as initial stage
40
41 snapshotStack.push(currentSnapshot);
42
43 // (Fifth rule)
44 while(!snapshotStack.empty())
45 {
46 currentSnapshotsnapshotStack.top();
47 snapshotStack.pop();
48
49 // (Sixth rule)
50 switch( currentSnapshot.stage)
51 {
52 case 0:
53 // (Seventh rule)
54 if( currentSnapshot.n0 )
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 } View Code 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 example2 int SomeFunc(int n, int retIdx)3 {4 ...5 if(n0)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 stackSnapShotStruct snapshotStack;
32 // (Fourth rule)
33 SnapShotStruct currentSnapshot;
34 currentSnapshot.n n; // set the value as parameter value
35 currentSnapshot.test0; // set the value as default value
36 currentSnapshot.stage0; // set the value as initial stage
37 snapshotStack.push(currentSnapshot);
38 // (Fifth rule)
39 while(!snapshotStack.empty())
40 {
41 currentSnapshotsnapshotStack.top();
42 snapshotStack.pop();
43 // (Sixth rule)
44 switch( currentSnapshot.stage)
45 {
46 case 0:
47 // (Seventh rule)
48 if( currentSnapshot.n0 )
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 } View Code 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 example2 int SomeFunc(int n, int retIdx)3 {4 ...5 if(n0)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 stackSnapShotStruct snapshotStack;
32 // (Fourth rule)
33 SnapShotStruct currentSnapshot;
34 currentSnapshot.n n; // set the value as parameter value
35 currentSnapshot.test0; // set the value as default value
36 currentSnapshot.stage0; // set the value as initial stage
37 snapshotStack.push(currentSnapshot);
38 // (Fifth rule)
39 while(!snapshotStack.empty())
40 {
41 currentSnapshotsnapshotStack.top();
42 snapshotStack.pop();
43 // (Sixth rule)
44 switch( currentSnapshot.stage)
45 {
46 case 0:
47 // (Seventh rule)
48 if( currentSnapshot.n0 )
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 } View Code 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 example2 int SomeFunc(int n, int retIdx)3 {4 ...5 if(n0)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 stackSnapShotStruct snapshotStack;
32 // (Fourth rule)
33 SnapShotStruct currentSnapshot;
34 currentSnapshot.n n; // set the value as parameter value
35 currentSnapshot.test0; // set the value as default value
36 currentSnapshot.stage0; // set the value as initial stage
37 snapshotStack.push(currentSnapshot);
38 // (Fifth rule)
39 while(!snapshotStack.empty())
40 {
41 currentSnapshotsnapshotStack.top();
42 snapshotStack.pop();
43 // (Sixth rule)
44 switch( currentSnapshot.stage)
45 {
46 case 0:
47 // (Seventh rule)
48 if( currentSnapshot.n0 )
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.test0; // - set the value as initial value
61 newSnapshot.stage0; // - 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 } View Code 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 stack5 using namespace std;6 7 /**8 * \brief 求n的阶乘9 * \para 10 * \return 11 * \note result n! 递归实现12 */13 int Fact(long n)14 {15 if(0n)16 return -1;17 if(0 n)18 return 1;19 else20 {21 return ( n* Fact(n-1));22 }23 } 24 25 /**26 * \brief 求n的阶乘27 * \para 28 * \return 29 * \note 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 stackSnapShotStruct snapshotStack;46 47 // (步骤4)48 SnapShotStruct currentSnapshot;49 currentSnapshot.inputNn;50 currentSnapshot.stage0; // 阶段变量初始化51 52 snapshotStack.push(currentSnapshot); 53 54 // (步骤5)55 while(!snapshotStack.empty()) 56 { 57 currentSnapshotsnapshotStack.top(); 58 snapshotStack.pop(); 59 60 // (步骤6)61 switch(currentSnapshot.stage)62 {63 // (步骤7)64 case 0:65 if(0currentSnapshot.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 else78 {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.stage1; // 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stage0 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__ View Code 2二分递归 1 #ifndef __BINARY_RECURSION_H__2 #define __BINARY_RECURSION_H__3 4 #include stack5 using namespace std;6 7 /**8 * \function FibNum9 * \brief 求斐波纳契数列10 * \para 11 * \return 12 * \note 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 FibNumLoop29 * \brief 求斐波纳契数列30 * \para 31 * \return 32 * \note 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 stackSnapShotStruct snapshotStack;50 51 // (步骤4)52 SnapShotStruct currentSnapshot;53 currentSnapshot.inputNn;54 currentSnapshot.stage0; // 阶段变量初始化55 56 snapshotStack.push(currentSnapshot);57 58 // (步骤5)59 while(!snapshotStack.empty())60 {61 currentSnapshotsnapshotStack.top();62 snapshotStack.pop();63 64 // (步骤6)65 switch(currentSnapshot.stage)66 {67 // (步骤7)68 case 0:69 if(currentSnapshot.inputN1)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 else82 {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.stage1; // 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.stage2; // 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__ View Code 3尾递归 1 #ifndef __TAIL_RECURSION_H__2 #define __TAIL_RECURSION_H__3 4 #include stack5 using namespace std;6 7 /**8 * \function FibNum29 * \brief 2阶裴波那契序列
10 * \para
11 * \return
12 * \note 递归实现 f0 x, f1 y, fnfn-1fn-2, nk,k1,...
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, xy);
23 }
24 }
25 /**
26 * \function FibNum2Loop
27 * \brief 2阶裴波那契序列
28 * \para
29 * \return
30 * \note 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 stackSnapShotStruct 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 currentSnapshotsnapshotStack.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, xy) )
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__ View Code 4互递归 1 #ifndef __MUTUAL_RECURSION_H__2 #define __MUTUAL_RECURSION_H__3 #include stack4 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 else18 // 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 else28 // 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 SnapShotStruct49 {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 stackSnapShotStruct 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 currentSnapshotsnapshotStack.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 else88 {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__ View Code 5嵌套递归 1 #ifndef __NESTED_RECURSION_H__2 #define __NESTED_RECURSION_H__3 #include stack4 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 else25 {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 stackSnapShotStruct 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 currentSnapshotsnapshotStack.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 method99 else
100 {
101 // (步骤10)
102
103 currentSnapshot.stage1;
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__ View Code 测试代码 1 #include tchar.h2 #include BinaryRecursion.h3 #include LinearRecursion.h4 #include MutualRecursion.h5 #include NestedRecursion.h6 #include TailRecursion.h7 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\n,result);
16 printf(FibNumLoop(10) %d\n,result2);
17
18
19 // Linear Recursion
20 result Fact(10);
21 result2 FactLoop(10);
22
23 printf(Fact(10) %d\n,result);
24 printf(FactLoop(10) %d\n,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\n,result);
32 printf(FibNumLoop2(10,5,4) %d\n,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\n,(int)bResult);
43 printf(IsOddNumberLoop(10) %d\n,(int)bResult2);
44 printf(IsEvenNumber(10) %d\n,(int)bResult3);
45 printf(IsEvenNumberLoop(10) %d\n,(int)bResult4);
46
47
48 // Nested Recursion
49 result Ackermann(3,2);
50 result2 AckermannLoop(3,2);
51
52 printf(Ackermann(3,2) %d\n,result);
53 printf(AckermannLoop(3,2) %d\n,result2);
54
55 while(1){}
56 return 0;
57 } View Code 6 更多的例子 epQuickSort.hepMergeSort.hepKAryHeap.hepPatriciaTree.h7 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 http://www.dreamincode.net/forums/topic/51296-types-of-recursion/EpLibrary 2.09 License 本文及包含的代码遵从协议 The MIT 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 转载于:https://www.cnblogs.com/wb-DarkHorse/p/3284228.html
