如果我们使用循环而不是 递归或者反之，在算法中两者都可以达到相同的目的?”

Usually yes if you are writing in a imperative language iteration will run faster than recursion, the performance hit is minimized in problems where the iterative solution requires manipulating Stacks and popping items off of a stack due to the recursive nature of the problem. There are a lot of times where the recursive implementation is much easier to read because the code is much shorter, so you do want to consider maintainability. Especailly in cases where the problem has a recursive nature. So take for example:

河内塔的递归实现:

def TowerOfHanoi(n , source, destination, auxiliary):
    if n==1:
        print ("Move disk 1 from source",source,"to destination",destination)
        return
    TowerOfHanoi(n-1, source, auxiliary, destination)
    print ("Move disk",n,"from source",source,"to destination",destination)
    TowerOfHanoi(n-1, auxiliary, destination, source)

相当短，很容易读。将其与对应的迭代TowerOfHanoi进行比较:

# Python3 program for iterative Tower of Hanoi
import sys
 
# A structure to represent a stack
class Stack:
    # Constructor to set the data of
    # the newly created tree node
    def __init__(self, capacity):
        self.capacity = capacity
        self.top = -1
        self.array = [0]*capacity
 
# function to create a stack of given capacity.
def createStack(capacity):
    stack = Stack(capacity)
    return stack
  
# Stack is full when top is equal to the last index
def isFull(stack):
    return (stack.top == (stack.capacity - 1))
   
# Stack is empty when top is equal to -1
def isEmpty(stack):
    return (stack.top == -1)
   
# Function to add an item to stack.
# It increases top by 1
def push(stack, item):
    if(isFull(stack)):
        return
    stack.top+=1
    stack.array[stack.top] = item
   
# Function to remove an item from stack.
# It decreases top by 1
def Pop(stack):
    if(isEmpty(stack)):
        return -sys.maxsize
    Top = stack.top
    stack.top-=1
    return stack.array[Top]
   
# Function to implement legal
# movement between two poles
def moveDisksBetweenTwoPoles(src, dest, s, d):
    pole1TopDisk = Pop(src)
    pole2TopDisk = Pop(dest)
 
    # When pole 1 is empty
    if (pole1TopDisk == -sys.maxsize):
        push(src, pole2TopDisk)
        moveDisk(d, s, pole2TopDisk)
       
    # When pole2 pole is empty
    else if (pole2TopDisk == -sys.maxsize):
        push(dest, pole1TopDisk)
        moveDisk(s, d, pole1TopDisk)
       
    # When top disk of pole1 > top disk of pole2
    else if (pole1TopDisk > pole2TopDisk):
        push(src, pole1TopDisk)
        push(src, pole2TopDisk)
        moveDisk(d, s, pole2TopDisk)
       
    # When top disk of pole1 < top disk of pole2
    else:
        push(dest, pole2TopDisk)
        push(dest, pole1TopDisk)
        moveDisk(s, d, pole1TopDisk)
   
# Function to show the movement of disks
def moveDisk(fromPeg, toPeg, disk):
    print("Move the disk", disk, "from '", fromPeg, "' to '", toPeg, "'")
   
# Function to implement TOH puzzle
def tohIterative(num_of_disks, src, aux, dest):
    s, d, a = 'S', 'D', 'A'
   
    # If number of disks is even, then interchange
    # destination pole and auxiliary pole
    if (num_of_disks % 2 == 0):
        temp = d
        d = a
        a = temp
    total_num_of_moves = int(pow(2, num_of_disks) - 1)
   
    # Larger disks will be pushed first
    for i in range(num_of_disks, 0, -1):
        push(src, i)
   
    for i in range(1, total_num_of_moves + 1):
        if (i % 3 == 1):
            moveDisksBetweenTwoPoles(src, dest, s, d)
   
        else if (i % 3 == 2):
            moveDisksBetweenTwoPoles(src, aux, s, a)
   
        else if (i % 3 == 0):
            moveDisksBetweenTwoPoles(aux, dest, a, d)
 
# Input: number of disks
num_of_disks = 3
 
# Create three stacks of size 'num_of_disks'
# to hold the disks
src = createStack(num_of_disks)
dest = createStack(num_of_disks)
aux = createStack(num_of_disks)
 
tohIterative(num_of_disks, src, aux, dest)

Now the first one is way easier to read because suprise suprise shorter code is usually easier to understand than code that is 10 times longer. Sometimes you want to ask yourself is the extra performance gain really worth it? The amount of hours wasted debugging the code. Is the iterative TowerOfHanoi faster than the Recursive TowerOfHanoi? Probably, but not by a big margin. Would I like to program Recursive problems like TowerOfHanoi using iteration? Hell no. Next we have another recursive function the Ackermann function: Using recursion:

    if m == 0:
        # BASE CASE
        return n + 1
    elif m > 0 and n == 0:
        # RECURSIVE CASE
        return ackermann(m - 1, 1)
    elif m > 0 and n > 0:
        # RECURSIVE CASE
        return ackermann(m - 1, ackermann(m, n - 1))

使用迭代:

callStack = [{'m': 2, 'n': 3, 'indentation': 0, 'instrPtr': 'start'}]
returnValue = None

while len(callStack) != 0:
    m = callStack[-1]['m']
    n = callStack[-1]['n']
    indentation = callStack[-1]['indentation']
    instrPtr = callStack[-1]['instrPtr']

    if instrPtr == 'start':
        print('%sackermann(%s, %s)' % (' ' * indentation, m, n))

        if m == 0:
            # BASE CASE
            returnValue = n + 1
            callStack.pop()
            continue
        elif m > 0 and n == 0:
            # RECURSIVE CASE
            callStack[-1]['instrPtr'] = 'after first recursive case'
            callStack.append({'m': m - 1, 'n': 1, 'indentation': indentation + 1, 'instrPtr': 'start'})
            continue
        elif m > 0 and n > 0:
            # RECURSIVE CASE
            callStack[-1]['instrPtr'] = 'after second recursive case, inner call'
            callStack.append({'m': m, 'n': n - 1, 'indentation': indentation + 1, 'instrPtr': 'start'})
            continue
    elif instrPtr == 'after first recursive case':
        returnValue = returnValue
        callStack.pop()
        continue
    elif instrPtr == 'after second recursive case, inner call':
        callStack[-1]['innerCallResult'] = returnValue
        callStack[-1]['instrPtr'] = 'after second recursive case, outer call'
        callStack.append({'m': m - 1, 'n': returnValue, 'indentation': indentation + 1, 'instrPtr': 'start'})
        continue
    elif instrPtr == 'after second recursive case, outer call':
        returnValue = returnValue
        callStack.pop()
        continue
print(returnValue)

再说一次，递归实现更容易理解。所以我的结论是，如果问题本质上是递归的，需要操作堆栈中的项，就使用递归。

使用Chrome 45.0.2454.85 m，递归似乎要快得多。

代码如下:

(function recursionVsForLoop(global) {
    "use strict";

    // Perf test
    function perfTest() {}

    perfTest.prototype.do = function(ns, fn) {
        console.time(ns);
        fn();
        console.timeEnd(ns);
    };

    // Recursion method
    (function recur() {
        var count = 0;
        global.recurFn = function recurFn(fn, cycles) {
            fn();
            count = count + 1;
            if (count !== cycles) recurFn(fn, cycles);
        };
    })();

    // Looped method
    function loopFn(fn, cycles) {
        for (var i = 0; i < cycles; i++) {
            fn();
        }
    }

    // Tests
    var curTest = new perfTest(),
        testsToRun = 100;

    curTest.do('recursion', function() {
        recurFn(function() {
            console.log('a recur run.');
        }, testsToRun);
    });

    curTest.do('loop', function() {
        loopFn(function() {
            console.log('a loop run.');
        }, testsToRun);
    });

})(window);

结果

//使用标准for循环运行100次

循环运行100x。 完成时间:7.683ms

//使用带有尾递归的函数递归方法运行100次

100x递归运行。 完成时间:4.841毫秒

在下面的截图中，当每次测试运行300次循环时，递归再次以更大的优势获胜

递归可能会更昂贵，这取决于递归函数是否是尾部递归(最后一行是递归调用)。尾递归应该被编译器识别，并优化为迭代的对应部分(同时保持代码中简洁、清晰的实现)。

我将以最有意义的方式编写算法，并且对那些不得不在几个月或几年内维护代码的可怜的傻瓜(无论是你自己还是其他人)来说是最清楚的。如果你遇到了性能问题，那就分析你的代码，然后，只有在那之后，你才能通过迭代实现来进行优化。您可能需要研究一下内存和动态编程。

你必须记住，使用太深的递归，你会遇到堆栈溢出，这取决于允许的堆栈大小。为了防止这种情况，请确保提供一些基本情况，以结束递归。

比较递归和迭代就像比较十字螺丝刀和一字螺丝刀。在大多数情况下，你可以拆卸任何一个平头的十字螺钉，但如果你使用专为该螺钉设计的螺丝刀，那就更容易了，对吧?

有些算法只是适合递归，因为它们的设计方式(斐波那契数列，遍历树状结构等)。递归使算法更简洁，更容易理解(因此可共享和可重用)。

此外，一些递归算法使用“惰性评估”，这使得它们比迭代算法更有效。这意味着它们只在需要的时候执行昂贵的计算，而不是每次循环运行时都执行。

这应该足够让你开始了。我也会给你找一些文章和例子。

链接1:Haskel vs PHP(递归vs迭代)

下面是一个程序员必须使用PHP处理大型数据集的示例。他展示了在Haskel中使用递归处理是多么容易，但由于PHP没有简单的方法来完成相同的方法，他被迫使用迭代来获得结果。

http://blog.webspecies.co.uk/2011-05-31/lazy-evaluation-with-php.html

链接2:掌握递归

递归的坏名声大多来自于命令式语言的高成本和低效率。本文的作者讨论了如何优化递归算法，使其更快、更有效。他还介绍了如何将传统循环转换为递归函数，以及使用尾部递归的好处。我认为他的结束语总结了我的一些要点:

递归编程为程序员提供了一种更好的组织方式 以一种既可维护又逻辑一致的方式编写代码。” https://developer.ibm.com/articles/l-recurs/

链接3:递归比循环快吗?(回答)

下面是一个与你的问题类似的stackoverflow问题的答案链接。作者指出，许多与递归或循环相关的基准测试都是特定于语言的。命令式语言通常使用循环更快，使用递归更慢，函数式语言反之亦然。我想从这个链接中得到的主要观点是，在语言不可知论/情境盲目的意义上回答这个问题是非常困难的。

递归比循环快吗?

这取决于语言。在Java中，你应该使用循环。函数式语言优化递归。