Javascript 有人能给我解释一下将这个递归函数添加到它本身是如何工作的吗?
这是我在想办法解决这个问题时遇到的 下面是函数(稍作修改以尝试帮助自己理解):Javascript 有人能给我解释一下将这个递归函数添加到它本身是如何工作的吗?,javascript,recursion,Javascript,Recursion,这是我在想办法解决这个问题时遇到的 下面是函数(稍作修改以尝试帮助自己理解): fibonacci(n-2)将被计算,然后紧接着fibonacci(n-1)。 但即使是这样,我也不明白JavaScript如何将这两个函数添加到一起 有谁能帮我理解一下这是怎么回事,或者至少,你能帮我以一种更容易理解的方式重新构造它吗 以下是输出: a brand new 20 consider yourself cached a brand new 18 consider yourself cached
fibonacci(n-2)
将被计算,然后紧接着fibonacci(n-1)
。但即使是这样,我也不明白JavaScript如何将这两个函数添加到一起 有谁能帮我理解一下这是怎么回事,或者至少,你能帮我以一种更容易理解的方式重新构造它吗 以下是输出:
a brand new 20 consider yourself cached
a brand new 18 consider yourself cached
a brand new 16 consider yourself cached
a brand new 14 consider yourself cached
a brand new 12 consider yourself cached
a brand new 10 consider yourself cached
a brand new 8 consider yourself cached
a brand new 6 consider yourself cached
a brand new 4 consider yourself cached
a brand new 2 consider yourself cached
no 0s or 1s, 0
no 0s or 1s, 1
current cache: Object {2: 1}
a brand new 3 consider yourself cached
no 0s or 1s, 1
already in the Object {2: 1}
current cache: Object {2: 1, 3: 2}
current cache: Object {2: 1, 3: 2, 4: 3}
a brand new 5 consider yourself cached
already in the Object {2: 1, 3: 2, 4: 3}
already in the Object {2: 1, 3: 2, 4: 3}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8}
a brand new 7 consider yourself cached
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8}
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21}
a brand new 9 consider yourself cached
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21}
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55}
a brand new 11 consider yourself cached
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55}
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144}
a brand new 13 consider yourself cached
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144}
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377}
a brand new 15 consider yourself cached
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377}
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377, 15: 610}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377, 15: 610, 16: 987}
a brand new 17 consider yourself cached
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377, 15: 610, 16: 987}
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377, 15: 610, 16: 987}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377, 15: 610, 16: 987, 17: 1597}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377, 15: 610, 16: 987, 17: 1597, 18: 2584}
a brand new 19 consider yourself cached
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377, 15: 610, 16: 987, 17: 1597, 18: 2584}
already in the Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377, 15: 610, 16: 987, 17: 1597, 18: 2584}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377, 15: 610, 16: 987, 17: 1597, 18: 2584, 19: 4181}
current cache: Object {2: 1, 3: 2, 4: 3, 5: 5, 6: 8, 7: 13, 8: 21, 9: 34, 10: 55, 11: 89, 12: 144, 13: 233, 14: 377, 15: 610, 16: 987, 17: 1597, 18: 2584, 19: 4181, 20: 6765}
谢谢,我知道递归可能是一个很大的noob问题,我已经用过好几次了,但了解它的工作原理让我头晕目眩。“我不明白JavaScript如何将这两个函数添加到一起。”
JS不添加函数,而是添加从这些函数调用返回的值。
假设f(n-2)已计算,当调用f(n-1)时,将通过以下公式计算:
f(n-1) = f(n-2) + f(n-3)
现在,由于我们在等式右侧计算了这两个值,所以这两个值都将从缓存中获取
让我们用一个例子来演示,假设我们要计算f(5):
具有f(3)的递归调用:
递归调用:
f(1) = 1 and f(2) = cached(f(1)) + f(0) = 1 + 0 = 1
现在我们回到计算f(3),我们缓存了f(2)和f(1),因此:
f(3) = 1 + 1 = 2
返回计算f(4)
再次结束:
f(5) = f(4) + f(3) = 3 + 2 = 5
请密切注意这样一个事实,即一旦到达递归的最深处(f(1)),就会一直使用缓存备份,并且不会计算任何值,这使得此实现非常高效 如果代码简化,并且从较小的数字开始,例如4,您可能会发现这更容易。以下功能与您最初发布的功能相同:
var cache = {};
function fibonacci(n) { // assume n = 4
var cached = cache[n];
第一次通过时,缓存[4]将未定义,因此以下测试的结果为false:
if (cached) {
console.log('already in the ', cache);
return cached;
}
当n=4时,以下也为假:
if (n <= 1) {
console.log('no 0s or 1s, ', n);
return n;
}
即:
cache[4] = fibonacci(2) + fibonacci(3);
在上行中,首先计算左侧,开始创建缓存的'4'属性。由于语句尚未完成,它实际上尚未创建,因此您几乎有:
然后计算右侧,以查看将分配什么。因为有一个+
运算符,所以必须对这两个表达式求值,以查看它是被视为加法还是级联
接下来对fibonacci(2)
进行求值(无论+
是加法还是缩合,求值从左到右进行),因此重复上述过程,创建:
cache[2] = fibonacci(0) + fibonacci(1);
你几乎做到了:
请注意,这些属性尚未实际创建
现在计算fibonacci(0)
。这次它到达第二个if
并返回0,因此不会创建缓存['0']
,现在您有:
cache[2] = 0 + fibonacci(1);
cache[3] = 1 + fibonacci(2);
cache[3] = 1 + 1; // cache = {3:1, 3:2, 4:undefined};
同样,在计算fibonacci(1)
时,第二个if
语句执行并返回1
,因此您有一个要分配的值,以便创建属性并分配值:
cache[2] = 0 + 1; // cache = {2:1, 4:undefined};
现在转到下一行:
console.log('current cache: ', cache);
return cache[n]; // returns 1;
}
现在,前面的通话继续:
cache[4] = 1 + fibonacci(3);
这又是一次:
cache[3] = fibonacci(1) + fibonacci(2);
第一个表达式从第一个if
返回1
,因此:
cache[2] = 0 + fibonacci(1);
cache[3] = 1 + fibonacci(2);
cache[3] = 1 + 1; // cache = {3:1, 3:2, 4:undefined};
第二个表达式到达第一个if,其中缓存[2]存在,因此它返回1(即缓存[2]的值),您有:
cache[2] = 0 + fibonacci(1);
cache[3] = 1 + fibonacci(2);
cache[3] = 1 + 1; // cache = {3:1, 3:2, 4:undefined};
然后返回缓存[3](即2),因此返回:
cache[4] = 1 + 2;
现在有cache={2:1,3:2,3:3}
您可以将上述内容作为顺序操作编写(所有递归函数都可以作为顺序操作编写),这在速度很重要的情况下很常见,因为顺序操作总是更快。在这种情况下,顺序函数非常简单:
function fibonacci2(n) {
var fibs = {0:0, 1:1};
var i = 1;
while (i < n) {
fibs[++i] = fibs[i-1] + fibs[i-2];
}
return fibs;
}
console.log(fibonacci2(4));
函数fibonacci2(n){
var fibs={0:0,1:1};
var i=1;
而(i
这是一个基本正确的答案-但是Javascript不会在f(3)
之前计算f(4)
在f(4)+f(3)
?@Patashu调用的顺序无关紧要,因为这个递归是自下而上计算的,所以f(3)总是在f(4)之前计算。令人困惑是的,有一点,但是试着像我那样用数字手动运行它,你会看到;)哦,是的,我把计算顺序和缓存顺序搞混了。非常感谢您的回答,我需要一点时间来消化它。请注意,您正在递归之前记录“认为自己已缓存”,因此这是实际缓存发生之前的方式。您可能需要使用Firebug的console.group()
功能来更好地记录日志。示例:非常感谢您的深入解释,因为它确实帮助我理解了输出。这将是几个小时之前,我得到测试出来,但我只是想表达我的赞赏马上!我认为这已经解决了,因为这正是我想要的,而且对我来说更容易理解。再次感谢您的帮助。经过一段时间的讨论,我想我的主要困惑是当n变为0时会发生什么。出于某种原因,我忘记了当函数返回时,函数只是将自身转换为返回值,这使得使用“+”运算符对两个函数的求值更容易理解。。。(即使递归仍然让我有点困惑)
cache[4] = 1 + 2;
function fibonacci2(n) {
var fibs = {0:0, 1:1};
var i = 1;
while (i < n) {
fibs[++i] = fibs[i-1] + fibs[i-2];
}
return fibs;
}
console.log(fibonacci2(4));