本文基于免费代码营基本算法脚本“分解数字”
在数学中,非负整数n的阶乘可能是一个棘手的算法。在本文中,我将解释这种方法,首先使用递归函数,第二种使用而循环,第三种使用以循环。
算法挑战
返回提供的整体的阶乘。
如果整体用字母n表示,则阶乘是所有小于或等于n的正整数的乘积。
阶乘经常用简写符号n!表示!
例如:5!= 1 * 2 * 3 * 4 * 5 = 120
function factorialize(num) {
return num;
}
factorialize(5);
提供的测试用例
- factorialize(0)应该返回1
- factorialize(5)应该返回120
- factorialize(10)应该返回3628800
- factorialize(20)应该返回2432902008176640000
什么是因数分解?
当将一个因数分解时,就是称为数字乘以每个连续的数字减一个。
如果您的电话号码是5,则您将:
5! = 5 * 4 * 3 * 2 * 1
该模式为:
0! = 1
1! = 1
2! = 2 * 1
3! = 3 * 2 * 1
4! = 4 * 3 * 2 * 1
5! = 5 * 4 * 3 * 2 * 1
1.递归分解一个数字
function factorialize(num) {
// If the number is less than 0, reject it.
if (num < 0)
return -1;
// If the number is 0, its factorial is 1.
else if (num == 0)
return 1;
// Otherwise, call the recursive procedure again
else {
return (num * factorialize(num - 1));
}
}
factorialize(5);
没有注释:
function factorialize(num) {
if (num < 0)
return -1;
else if (num == 0)
return 1;
else {
return (num * factorialize(num - 1));
}
}
factorialize(5);
2.用WHILE循环分解一个数字
function factorialize(num) {
// Step 1. Create a variable result to store num
var result = num;
// If num = 0 OR num = 1, the factorial will return 1
if (num === 0 || num === 1)
return 1;
// Step 2. Create the WHILE loop
while (num > 1) {
num--; // decrementation by 1 at each iteration
result = result * num; // or result *= num;
}
// Step 3. Return the factorial of the provided integer
return result; // 120
}
factorialize(5);
没有注释:
function factorialize(num) {
var result = num;
if (num === 0 || num === 1)
return 1;
while (num > 1) {
num--;
result *= num;
}
return result;
}
factorialize(5);
3.使用FOR循环分解数字
function factorialize(num) {
// If num = 0 OR num = 1, the factorial will return 1
if (num === 0 || num === 1)
return 1;
// We start the FOR loop with i = 4
// We decrement i after each iteration
for (var i = num - 1; i >= 1; i--) {
// We store the value of num at each iteration
num = num * i; // or num *= i;
}
return num; //120
}
factorialize(5);
没有注释:
function factorialize(num) {
if (num === 0 || num === 1)
return 1;
for (var i = num - 1; i >= 1; i--) {
num *= i;
}
return num;
}
factorialize(5);
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