How to Merge sorted Arrays in JavaScript

Question

I have three sorted arrays like below

[{name:"a"}, {name:"b"}, {name:"m"}, {name:"x"}]
[{name:"a"}, {name:"e"}, {name:"i"}, {name:"o"}]
[{name:"g"}, {name:"h"}, {name:"m"}, {name:"n"}]

Those arrays are sorted based on name property of each object in Array. Here is the method I converted from Java to merge two sorted arrays

function mergeSorted(a, b) {
  var answer = new Array(a.length + b.length), i = 0, j = 0, k = 0;
  while (i < a.length && j < b.length) {
    if (a[i].name < b[j].name) {
        answer[k] = a[i];
        i++;
    }else {
        answer[k] = b[j];
        j++;
    }
    k++;
  }
  while (i < a.length) {
    answer[k] = a[i];
    i++;
    k++;
  }
  while (j < b.length) {
    answer[k] = b[j];
    j++;
    k++;
  }
  return answer;
}

Here is the working fiddle with two arrays http://jsfiddle.net/euRn5/. What is the best approach to achieve the same with n number of Arrays, the thought I have in my mind currently is take one by one, merge it with previously merged till the last item, like n += i stuff. Is this a best approach?

Answer 1

The native implementations are not always the fastest (as you may have noticed) and have, historically, been somewhat sluggish, due to extensive error checking. That being said, there may be performance enhancements in the future, due to more robust integration with the hardware or routines specifically built to optimize certain tasks. If you write your own code, your application won't be able to take advantage of these boosts in performance once they're implemented. It's up to you to decide where the advantages lie and what the risks are.

At any rate, I've written a prettier version of your optimized code for funsies:

function mergeSorted(a,b){
    var alen = a.length
      , blen = b.length
      , i, j, k = j = i = 0
      , answer = new Array(alen + blen)
    ;//var

    while(i < alen && j < blen)
                    answer[k++] = a[i].name < b[j].name ? a[i++] : b[j++];
    while(i < alen) answer[k++] = a[i++];
    while(j < blen) answer[k++] = b[j++];

    return answer;
}