Recursive call (---Fibonacci sequence)

Recursive: Call yourself again inside a function;

Efficiency: This call has not yet ended , the next call will start, and this call will be suspended until all calls are completed before returning in sequence.

It is shaped like a mathematical function: the factorial of f(n)

##ex: 5!=5*4! (5* 4*3*2*1)

4!=4*3! (4*3*2*1)

3!=3*2! (3*2*1)

2!=2*1! ( 2*1)

##1!=1;

F(n)! =n*f(n-1)!

<script>
	//计算数字n的阶乘 （if方法）
		function f(n){
			if(n==1){            //边界条件
				return 1;
			}else{               //没到边界条件
				return n*f(n-1);
			}
		}

	/*或者：return n==1 ? 1 : n*f(n-1); （三目运算方法）
		function f(n){
			var result = return n==1 ? 1 : n*f(n-1);
			return result;
		}
	*/
		//计算5的阶乘？
		function testF(){
			var result = f(5);
			console.log(result);
		}
		testF();
	</script>

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Exercise:

The following sequence: Fibonacci sequence
1,1,2,3,5,8,13 ,21,34,55………….
Known: The first number and the second number in the sequence are both 1
From the third Starting with numbers, each number is the sum of the previous two numbers

Question: Use a function to find the nth number in the sequence The value of

<!doctype html>
<html lang="en">
 <head>
	 <meta charset="UTF-8">
     <title>Document</title>
	 <link rel="stylesheet" style="text/css" href="">
	 <style>
	 </style>
 </head>
 <body>
	<script>
	 function f(n){
			if(n==1 || n==2){
			  return 1;
			 }else{
			  return f(n-1)+f(n-2);
			 }
		}
		function testF(){
		var result=f(20);
		console.log(result);
		}
		testF();
	</script>	
 </body>
</html>

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