javascript递归string1的第一个字符，然后string2的最后一个字符

Question

Suppose I have two strings, n1 = abc and n2 = xyz. How can I print "azbycx" recursively?

Here is my code. I am practicing string manipulations and I can't seem to find a solution to this. I Googled already and can't seem to find one.

document.write(practice("abc","xyz"));

function practice(n1,n2){

if(n1.length==0){return n2;}
if(n2.length==0){return n1;}

return n1.substring(0,1) + practice(n2.charAt(n2.length-1),n1.substring(1));
}

It prints only "azc" :(

Answer 1

 function practice(n1,n2){
    console.log(n1);
    console.log(n2);
    if(n1.length==0){return n2;}
    if(n2.length==0){return n1;}


    return n1[0]+n2[n2.length-1] + practice(n1.substring(1), n2.substring(0, n2.length-1));
}

Answer 2

function practice(n1,n2){

    if(n1.length==0){return n2;}
    if(n2.length==0){return n1;}

    if (n2.length > n1.length) {
        return n2.charAt(n2.length-1) + practice(n1, n2.substring(0, n2.length-1));
    }
    else {
        return n1.charAt(0) + practice(n1.substring(1), n2);
    }
}

Answer 3

var practice = function(n1,n2) {
    var _charA = n1.charAt(0)
    var _charB = n2.charAt(n2.length-1)

    var _res = "";
    if(_charA != "" || _charB != "") {
        _res = practice(n1.substring(1,[n1.length]),n2.substring(0,[n2.length-1]))
    }
    return _charA + _charB + _res;
}

practice("abcd", "uvwxyz")
//output - azbycxdwvu

Answer 4

This would do what you like it to do:

function practice(n1,n2){
  if(n1.length==0 && n2.length==0)
    return "";

  return n1.substr(0,1) + n2.substr(n2.length-1, 1) + 
         practice(n1.substr(1, n1.length-1), n2.substr(0, n2.length-1));
}

jsfiddle here: http://jsfiddle.net/gt7fsb7g/