Javascript中的排序算法 [英] Sorting algorithm in Javascript

问题描述

为jQuery函数$("#sort").click写一个JavaScript回调.允许用户以任何顺序输入三个数字.按从低到高的顺序输出数字.

Write a JavaScript callback for the jQuery function $("#sort").click. Allow the user to enter three numbers in any order. Output the numbers in order from lowest to highest.

$(document).ready(function() {
  $("#sort").click(function() {
    var a = Number($("#a").val());
    var b = Number($("#b").val());
    var c = Number($("#c").val());
    var message = "";
   if (b > c) {
     if ((b + c) > (a + c)) {
       message = c + " " + a + " " + b;
     } else {
       message = c + " " + b + " " + a;
     }
   } else {
     message = b + " " + a + " " + c;
   }
    if (b > a) {
      if ((a + b) > (a + c)) {
        message = a + " " + c + " " + b;
      } else {
        message = a + " " + b + " " + c;
      }
    } else {
      message = b + " " + c + " " + a;
    }
  $("#output").html(message)
  });
});

任何人都介意看一下这段代码并说出什么问题吗?

Would anyone mind looking at this code and saying what's wrong?

推荐答案

这里有一些很棒的jQuery答案，我将介绍比较部分.

Some great jQuery answers here, I'll cover the comparisons part.

您不需要进行五次比较，只需进行三次比较(如果幸运的话就可以进行两次).比较a和b，并交换a > b.然后比较c和b.如果为c > b，则说明已完成，否则请比较c和a:

You don't need five comparisons, just three (or two if you're lucky). Compare a and b, and swap them if a > b. Then compare c and b. If c > b, you're done, otherwise compare c and a:

if (a > b)
  x = a, a = b, b = x;

if (c < b)
  if (c < a)
    result = [c, a, b];
  else
    result = [a, c, b];
else
  result = [a, b, c];

如果所有数字均为32位正整数，则可以对它们进行排序，而无需进行任何比较:

If all numbers are 32-bit positive ints, you can sort them without any comparisons at all:

min = function(a,b) { return b + ((a-b) & ((a-b)>>31)) }
max = function(a,b) { return a - ((a-b) & ((a-b)>>31)) }

x = min(a, min(b, c));
z = max(a, max(b, c));
y = (a + b + c) - (x + z);

result = [x, y, z];

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