问题
I've used recursion quite a lot on my many years of programming to solve simple problems, but I'm fully aware that sometimes you need iteration due to memory/speed problems.
So, sometime in the very far past I went to try and find if there existed any "pattern" or text-book way of transforming a common recursion approach to iteration and found nothing. Or at least nothing that I can remember it would help.
- Are there general rules?
- Is there a "pattern"?
回答1:
Usually, I replace a recursive algorithm by an iterative algorithm by pushing the parameters that would normally be passed to the recursive function onto a stack. In fact, you are replacing the program stack by one of your own.
Stack<Object> stack;
stack.push(first_object);
while( !stack.isEmpty() ) {
// Do something
my_object = stack.pop();
// Push other objects on the stack.
}
Note: if you have more than one recursive call inside and you want to preserve the order of the calls, you have to add them in the reverse order to the stack:
foo(first);
foo(second);
has to be replaced by
stack.push(second);
stack.push(first);
Edit: The article Stacks and Recursion Elimination (or Article Backup link) goes into more details on this subject.
回答2:
Really, the most common way to do it is to keep your own stack. Here's a recursive quicksort function in C:
void quicksort(int* array, int left, int right)
{
if(left >= right)
return;
int index = partition(array, left, right);
quicksort(array, left, index - 1);
quicksort(array, index + 1, right);
}
Here's how we could make it iterative by keeping our own stack:
void quicksort(int *array, int left, int right)
{
int stack[1024];
int i=0;
stack[i++] = left;
stack[i++] = right;
while (i > 0)
{
right = stack[--i];
left = stack[--i];
if (left >= right)
continue;
int index = partition(array, left, right);
stack[i++] = left;
stack[i++] = index - 1;
stack[i++] = index + 1;
stack[i++] = right;
}
}
Obviously, this example doesn't check stack boundaries... and really you could size the stack based on the worst case given left and and right values. But you get the idea.
回答3:
It seems nobody has addressed where the recursive function calls itself more than once in the body, and handles returning to a specific point in the recursion (i.e. not primitive-recursive). It is said that every recursion can be turned into iteration, so it appears that this should be possible.
I just came up with a C# example of how to do this. Suppose you have the following recursive function, which acts like a postorder traversal, and that AbcTreeNode is a 3-ary tree with pointers a, b, c.
public static void AbcRecursiveTraversal(this AbcTreeNode x, List<int> list) {
if (x != null) {
AbcRecursiveTraversal(x.a, list);
AbcRecursiveTraversal(x.b, list);
AbcRecursiveTraversal(x.c, list);
list.Add(x.key);//finally visit root
}
}
The iterative solution:
int? address = null;
AbcTreeNode x = null;
x = root;
address = A;
stack.Push(x);
stack.Push(null)
while (stack.Count > 0) {
bool @return = x == null;
if (@return == false) {
switch (address) {
case A://
stack.Push(x);
stack.Push(B);
x = x.a;
address = A;
break;
case B:
stack.Push(x);
stack.Push(C);
x = x.b;
address = A;
break;
case C:
stack.Push(x);
stack.Push(null);
x = x.c;
address = A;
break;
case null:
list_iterative.Add(x.key);
@return = true;
break;
}
}
if (@return == true) {
address = (int?)stack.Pop();
x = (AbcTreeNode)stack.Pop();
}
}
回答4:
Strive to make your recursive call Tail Recursion (recursion where the last statement is the recursive call). Once you have that, converting it to iteration is generally pretty easy.
回答5:
Well, in general, recursion can be mimicked as iteration by simply using a storage variable. Note that recursion and iteration are generally equivalent; one can almost always be converted to the other. A tail-recursive function is very easily converted to an iterative one. Just make the accumulator variable a local one, and iterate instead of recurse. Here's an example in C++ (C were it not for the use of a default argument):
// tail-recursive
int factorial (int n, int acc = 1)
{
if (n == 1)
return acc;
else
return factorial(n - 1, acc * n);
}
// iterative
int factorial (int n)
{
int acc = 1;
for (; n > 1; --n)
acc *= n;
return acc;
}
Knowing me, I probably made a mistake in the code, but the idea is there.
回答6:
Even using stack will not convert a recursive algorithm into iterative. Normal recursion is function based recursion and if we use stack then it becomes stack based recursion. But its still recursion.
For recursive algorithms, space complexity is O(N) and time complexity is O(N). For iterative algorithms, space complexity is O(1) and time complexity is O(N).
But if we use stack things in terms of complexity remains same. I think only tail recursion can be converted into iteration.
回答7:
The stacks and recursion elimination article captures the idea of externalizing the stack frame on heap, but does not provide a straightforward and repeatable way to convert. Below is one.
While converting to iterative code, one must be aware that the recursive call may happen from an arbitrarily deep code block. Its not just the parameters, but also the point to return to the logic that remains to be executed and the state of variables which participate in subsequent conditionals, which matter. Below is a very simple way to convert to iterative code with least changes.
Consider this recursive code:
struct tnode
{
tnode(int n) : data(n), left(0), right(0) {}
tnode *left, *right;
int data;
};
void insertnode_recur(tnode *node, int num)
{
if(node->data <= num)
{
if(node->right == NULL)
node->right = new tnode(num);
else
insertnode(node->right, num);
}
else
{
if(node->left == NULL)
node->left = new tnode(num);
else
insertnode(node->left, num);
}
}
Iterative code:
// Identify the stack variables that need to be preserved across stack
// invocations, that is, across iterations and wrap them in an object
struct stackitem
{
stackitem(tnode *t, int n) : node(t), num(n), ra(0) {}
tnode *node; int num;
int ra; //to point of return
};
void insertnode_iter(tnode *node, int num)
{
vector<stackitem> v;
//pushing a stackitem is equivalent to making a recursive call.
v.push_back(stackitem(node, num));
while(v.size())
{
// taking a modifiable reference to the stack item makes prepending
// 'si.' to auto variables in recursive logic suffice
// e.g., instead of num, replace with si.num.
stackitem &si = v.back();
switch(si.ra)
{
// this jump simulates resuming execution after return from recursive
// call
case 1: goto ra1;
case 2: goto ra2;
default: break;
}
if(si.node->data <= si.num)
{
if(si.node->right == NULL)
si.node->right = new tnode(si.num);
else
{
// replace a recursive call with below statements
// (a) save return point,
// (b) push stack item with new stackitem,
// (c) continue statement to make loop pick up and start
// processing new stack item,
// (d) a return point label
// (e) optional semi-colon, if resume point is an end
// of a block.
si.ra=1;
v.push_back(stackitem(si.node->right, si.num));
continue;
ra1: ;
}
}
else
{
if(si.node->left == NULL)
si.node->left = new tnode(si.num);
else
{
si.ra=2;
v.push_back(stackitem(si.node->left, si.num));
continue;
ra2: ;
}
}
v.pop_back();
}
}
Notice how the structure of the code still remains true to the recursive logic and modifications are minimal, resulting in less number of bugs. For comparison, I have marked the changes with ++ and --. Most of the new inserted blocks except v.push_back, are common to any converted iterative logic
void insertnode_iter(tnode *node, int num)
{
+++++++++++++++++++++++++
vector<stackitem> v;
v.push_back(stackitem(node, num));
while(v.size())
{
stackitem &si = v.back();
switch(si.ra)
{
case 1: goto ra1;
case 2: goto ra2;
default: break;
}
------------------------
if(si.node->data <= si.num)
{
if(si.node->right == NULL)
si.node->right = new tnode(si.num);
else
{
+++++++++++++++++++++++++
si.ra=1;
v.push_back(stackitem(si.node->right, si.num));
continue;
ra1: ;
-------------------------
}
}
else
{
if(si.node->left == NULL)
si.node->left = new tnode(si.num);
else
{
+++++++++++++++++++++++++
si.ra=2;
v.push_back(stackitem(si.node->left, si.num));
continue;
ra2: ;
-------------------------
}
}
+++++++++++++++++++++++++
v.pop_back();
}
-------------------------
}
回答8:
Search google for "Continuation passing style." There is a general procedure for converting to a tail recursive style; there is also a general procedure for turning tail recursive functions into loops.
回答9:
Just killing time... A recursive function
void foo(Node* node)
{
if(node == NULL)
return;
// Do something with node...
foo(node->left);
foo(node->right);
}
can be converted to
void foo(Node* node)
{
if(node == NULL)
return;
// Do something with node...
stack.push(node->right);
stack.push(node->left);
while(!stack.empty()) {
node1 = stack.pop();
if(node1 == NULL)
continue;
// Do something with node1...
stack.push(node1->right);
stack.push(node1->left);
}
}
回答10:
Generally the technique to avoid stack overflow is for recursive functions is called trampoline technique which is widely adopted by Java devs.
However, for C# there is a little helper method here that turns your recursive function to iterative without requiring to change logic or make the code in-comprehensible. C# is such a nice language that amazing stuff is possible with it.
It works by wrapping parts of the method by a helper method. For example the following recursive function:
int Sum(int index, int[] array)
{
//This is the termination condition
if (int >= array.Length)
//This is the returning value when termination condition is true
return 0;
//This is the recursive call
var sumofrest = Sum(index+1, array);
//This is the work to do with the current item and the
//result of recursive call
return array[index]+sumofrest;
}
Turns into:
int Sum(int[] ar)
{
return RecursionHelper<int>.CreateSingular(i => i >= ar.Length, i => 0)
.RecursiveCall((i, rv) => i + 1)
.Do((i, rv) => ar[i] + rv)
.Execute(0);
}
回答11:
Thinking of things that actually need a stack:
If we consider the pattern of recursion as:
if(task can be done directly) {
return result of doing task directly
} else {
split task into two or more parts
solve for each part (possibly by recursing)
return result constructed by combining these solutions
}
For example, the classic Tower of Hanoi
if(the number of discs to move is 1) {
just move it
} else {
move n-1 discs to the spare peg
move the remaining disc to the target peg
move n-1 discs from the spare peg to the target peg, using the current peg as a spare
}
This can be translated into a loop working on an explicit stack, by restating it as:
place seed task on stack
while stack is not empty
take a task off the stack
if(task can be done directly) {
Do it
} else {
Split task into two or more parts
Place task to consolidate results on stack
Place each task on stack
}
}
For Tower of Hanoi this becomes:
stack.push(new Task(size, from, to, spare));
while(! stack.isEmpty()) {
task = stack.pop();
if(task.size() = 1) {
just move it
} else {
stack.push(new Task(task.size() -1, task.spare(), task,to(), task,from()));
stack.push(new Task(1, task.from(), task.to(), task.spare()));
stack.push(new Task(task.size() -1, task.from(), task.spare(), task.to()));
}
}
There is considerable flexibility here as to how you define your stack. You can make your stack a list of Command objects that do sophisticated things. Or you can go the opposite direction and make it a list of simpler types (e.g. a "task" might be 4 elements on a stack of int, rather than one element on a stack of Task).
All this means is that the memory for the stack is in the heap rather than in the Java execution stack, but this can be useful in that you have more control over it.
回答12:
One pattern to look for is a recursion call at the end of the function (so called tail-recursion). This can easily be replaced with a while. For example, the function foo:
void foo(Node* node)
{
if(node == NULL)
return;
// Do something with node...
foo(node->left);
foo(node->right);
}
ends with a call to foo. This can be replaced with:
void foo(Node* node)
{
while(node != NULL)
{
// Do something with node...
foo(node->left);
node = node->right;
}
}
which eliminates the second recursive call.
回答13:
A question that had been closed as a duplicate of this one had a very specific data structure:
The node had the following structure:
typedef struct {
int32_t type;
int32_t valueint;
double valuedouble;
struct cNODE *next;
struct cNODE *prev;
struct cNODE *child;
} cNODE;
The recursive deletion function looked like:
void cNODE_Delete(cNODE *c) {
cNODE*next;
while (c) {
next=c->next;
if (c->child) {
cNODE_Delete(c->child)
}
free(c);
c=next;
}
}
In general, it is not always possible to avoid a stack for recursive functions that invoke itself more than one time (or even once). However, for this particular structure, it is possible. The idea is to flatten all the nodes into a single list. This is accomplished by putting the current node's child at the end of the top row's list.
void cNODE_Delete (cNODE *c) {
cNODE *tmp, *last = c;
while (c) {
while (last->next) {
last = last->next; /* find last */
}
if ((tmp = c->child)) {
c->child = NULL; /* append child to last */
last->next = tmp;
tmp->prev = last;
}
tmp = c->next; /* remove current */
free(c);
c = tmp;
}
}
This technique can be applied to any data linked structure that can be reduce to a DAG with a deterministic topological ordering. The current nodes children are rearranged so that the last child adopts all of the other children. Then the current node can be deleted and traversal can then iterate to the remaining child.
回答14:
Recursion is nothing but the process of calling of one function from the other only this process is done by calling of a function by itself. As we know when one function calls the other function the first function saves its state(its variables) and then passes the control to the called function. The called function can be called by using the same name of variables ex fun1(a) can call fun2(a). When we do recursive call nothing new happens. One function calls itself by passing the same type and similar in name variables(but obviously the values stored in variables are different,only the name remains same.)to itself. But before every call the function saves its state and this process of saving continues. The SAVING IS DONE ON A STACK.
NOW THE STACK COMES INTO PLAY.
So if you write an iterative program and save the state on a stack each time and then pop out the values from stack when needed, you have successfully converted a recursive program into an iterative one!
The proof is simple and analytical.
In recursion the computer maintains a stack and in iterative version you will have to manually maintain the stack.
Think over it, just convert a depth first search(on graphs) recursive program into a dfs iterative program.
All the best!
回答15:
Another simple and complete example of turning the recursive function into iterative one using the stack.
#include <iostream>
#include <stack>
using namespace std;
int GCD(int a, int b) { return b == 0 ? a : GCD(b, a % b); }
struct Par
{
int a, b;
Par() : Par(0, 0) {}
Par(int _a, int _b) : a(_a), b(_b) {}
};
int GCDIter(int a, int b)
{
stack<Par> rcstack;
if (b == 0)
return a;
rcstack.push(Par(b, a % b));
Par p;
while (!rcstack.empty())
{
p = rcstack.top();
rcstack.pop();
if (p.b == 0)
continue;
rcstack.push(Par(p.b, p.a % p.b));
}
return p.a;
}
int main()
{
//cout << GCD(24, 36) << endl;
cout << GCDIter(81, 36) << endl;
cin.get();
return 0;
}
回答16:
A rough description of how a system takes any recursive function and executes it using a stack:
This intended to show the idea without details. Consider this function that would print out nodes of a graph:
function show(node)
0. if isleaf(node):
1. print node.name
2. else:
3. show(node.left)
4. show(node)
5. show(node.right)
For example graph: A->B A->C show(A) would print B, A, C
Function calls mean save the local state and the continuation point so you can come back, and then jump the the function you want to call.
For example, suppose show(A) begins to run. The function call on line 3. show(B) means - Add item to the stack meaning "you'll need to continue at line 2 with local variable state node=A" - Goto line 0 with node=B.
To execute code, the system runs through the instructions. When a function call is encountered, the system pushes information it needs to come back to where it was, runs the function code, and when the function completes, pops the information about where it needs to go to continue.
回答17:
This link provides some explanation and proposes the idea of keeping "location" to be able to get to the exact place between several recursive calls:
However, all these examples describe scenarios in which a recursive call is made a fixed amount of times. Things get trickier when you have something like:
function rec(...) {
for/while loop {
var x = rec(...)
// make a side effect involving return value x
}
}
回答18:
There is a general way of converting recursive traversal to iterator by using a lazy iterator which concatenates multiple iterator suppliers (lambda expression which returns an iterator). See my Converting Recursive Traversal to Iterator.
来源：https://stackoverflow.com/questions/38263100/deleting-linked-list-nodes-without-recursion