Detailed explanation of the usage of JS recursion

recursion:

The function calls the function itself, which is recursion. Recursion must have an end condition

    function f1() {
        console.log("从前有座山，山里有个庙，庙里有个老和尚给小和尚讲故事：");
        f1();
    };
    f1();//浏览器崩溃，因为没有结束条件——死循环

	改进如下：
	    var i=0;
    function f1() {
        i++;
        if (i<5){
            f1();
        }
        console.log("从前有座山，山里有个庙，庙里有个老和尚给小和尚讲故事：");
    };
    f1();

Related learning tutorial: javascript tutorial

Little Lizi:

Recursive implementation: find the sum of n numbers n=5 ------->5 4 3 2 1

//for 循环写法：
    var sum=0;
    for (var i=0;i<=5;i++){
        sum+=i;
    }
    console.log(sum);
----------------------分割线---------------------------

   function getSum(x) {
        if (x==1){
          return 1
        }
        return x+getSum(x-1);
    };

    var sum1=getSum(5);
    console.log(sum1);
    console.log(getSum(10));

Execution process:
The code executes getSum(5)—>Enter the function, x is 5 at this time, and 5 getSum(4) is executed. This When the code waits
At this time 5 getSum(4), the code does not calculate first, executes getSum(4) first, enters the function, executes 4 getSum(3), wait, executes getSum(3) first, Enter the function, execute 3 getSum(2), wait, execute getSum(2) first, enter the function, execute 2 getSum(1); wait, execute getSum(1) first, execute the judgment of x==1, return 1 ,So,
The result of getSum(1) at this time is 1, start to go out
2 getSum(1) The result at this time is: 2 1
Execution:
getSum(2) ---->2 1
3 getSum(2) The result at this time is 3 2 1
4 getSum(3) The result at this time is 4 3 2 1
5 getSum(4) This The result is 5 4 3 2 1

    结果:15

A few more:

    //递归案例:求一个数字各个位数上的数字的和:  123   --->6 ---1+2+3
    //523
    function getEverySum(x) {
        if(x<10){
            return x;
        }
        //获取的是这个数字的个位数
        return x%10+getEverySum(parseInt(x/10));
    }
    console.log(getEverySum(1364));//5

 //递归案例:求斐波那契数列

    function getFib(x) {
        if(x==1||x==2){
            return 1
        }
        return getFib(x-1)+getFib(x-2);
    }
    console.log(getFib(12));

Recursion:

The function calls the function itself, which is recursion. The recursion must have an end condition

    function f1() {
        console.log("从前有座山，山里有个庙，庙里有个老和尚给小和尚讲故事：");
        f1();
    };
    f1();//浏览器崩溃，因为没有结束条件——死循环

	改进如下：
	    var i=0;
    function f1() {
        i++;
        if (i<5){
            f1();
        }
        console.log("从前有座山，山里有个庙，庙里有个老和尚给小和尚讲故事：");
    };
    f1();

Little chestnuts:

Recursive implementation: Find the sum of n numbers n=5 ------->5 4 3 2 1

//for 循环写法：
    var sum=0;
    for (var i=0;i<=5;i++){
        sum+=i;
    }
    console.log(sum);
----------------------分割线---------------------------

   function getSum(x) {
        if (x==1){
          return 1
        }
        return x+getSum(x-1);
    };

    var sum1=getSum(5);
    console.log(sum1);
    console.log(getSum(10));

Execution process:
The code executes getSum(5)—>Enter the function. At this time, x is 5, and 5 getSum(4) is executed. At this time, the code waits.
At this time, 5 getSum(4), the code Do not calculate first, execute getSum(4) first, enter the function, execute 4 getSum(3), wait, execute getSum(3) first, enter the function, execute 3 getSum(2), wait, execute getSum first (2), enter the function, execute 2 getSum(1); wait, execute getSum(1) first, execute the judgment of x==1, return 1, so,
The result of getSum(1) at this time is 1. Start walking out
2 getSum(1) The result at this time is: 2 1
Execution:
getSum(2)---->2 1
3 getSum(2 ) The result at this time is 3 2 1
4 getSum(3) The result at this time is 4 3 2 1
5 getSum(4) The result at this time is 5 4 3 2 1

    结果:15

A few more:

    //递归案例:求一个数字各个位数上的数字的和:  123   --->6 ---1+2+3
    //523
    function getEverySum(x) {
        if(x<10){
            return x;
        }
        //获取的是这个数字的个位数
        return x%10+getEverySum(parseInt(x/10));
    }
    console.log(getEverySum(1364));//5

rrree

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