排序根据每个元件的长度的阵列 [英] Sort an array based on the length of each element

问题描述

我有一个这样的数组：

arr = []
arr[0] = "ab"
arr[1] = "abcdefgh"
arr[2] = "abcd"

排序后，输出阵列应该是：

After sorting, the output array should be:

arr[0] = "abcdefgh"
arr[1] = "abcd"
arr[2] = "ab"  

我的意思是，我想在各元件的长度的递减顺序

I mean, I want in the descending order of the length of each element.

推荐答案

您可以使用的 中的Array.sort 方法数​​组排序。排序函数认为作为排序标准可用于串的长度，如下所示：

You can use Array.sort method to sort the array. A sorting function that considers the length of string as the sorting criteria can be used as follows:

arr.sort(function(a, b){
  // ASC  -> a.length - b.length
  // DESC -> b.length - a.length
  return b.length - a.length;
});

---

注：排序 [A，B，C] 是字符串的长度不保证返回 [一，b，C] 。据规格的：

---

Note: sorting ["a", "b", "c"] by length of string is not guaranteed to return ["a", "b", "c"]. According to the specs:

的排序是未必稳定（即，比较元素
  平等不一定留在原来的顺序）。

The sort is not necessarily stable (that is, elements that compare equal do not necessarily remain in their original order).

如果目标是按长度排序然后用字典顺序必须指定附加标准：

If the objective is to sort by length then by dictionary order you must specify additional criteria:

["c", "a", "b"].sort(function(a, b) {
  return a.length - b.length || // sort by length, if equal then
         a.localeCompare(b);    // sort by dictionary order
});

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