问在给定两个n元组的情况下，如何有效地实现/构造以下排列？

EN

function qperm (x, y, xprime, yprime) {
  var i;
  var n = x.length;
  var qperm = new Array(n);
  var countsx = [], countsy = []; // same as new Array()
  qperm[xprime] = yprime; // doing 1.

  for (i = 0; i < n; ++i) {
    if (x[i] == y[i] && i != xprime && i != yprime) { // doing 2.
      qperm[i] = i; }
    else { // preparing for 4. below
      if (i != xprime) {
        if (countsx[x[i]]) countsx[x[i]]++; else countsx[x[i]] = 1; }
      if (i != yprime) {
        if (countsy[y[i]]) countsy[y[i]]++; else countsy[y[i]] = 1; } }

  // finishing countsx and countsy
  var count, sum;
  for (i = 0, count = 0; i < countsx.length; ++i) {
    if (countsx[i]) {
      sum = count + countsx[i];
      countsx[i] = count;
      count = sum; }
  for (i = 0, count = 0; i < countsy.length; ++i) {
    if (countsy[i]) {
      sum = count + countsy[i];
      countsy[i] = count;
      count = sum; }

  var yranked = new Array(count);      
  for (i = 0; i < n; ++i) {
    if (i != yprime && (x[i] != y[i] || i == xprime)) { // doing 4. for y
      yranked[countsy[y[i]]] = y[i];
      countsy[y[i]]++; } }

  for (i = 0; i < n; ++i) {
    if (i != xprime && (x[i] != y[i] || i == yprime)) { // doing 4. for x and 5. at the same time
      // this was here but was not right: qperm[x[i]] = yranked[countsx[x[i]]];
      qperm[i] = yranked[countsx[x[i]]];
      // this was here but was not right: countsy[y[i]]++; } } }
      countsx[x[i]]++; } }

  return qperm; }