循环创建相同长度字符串的数组

Posted 2023-03-15

【中文标题】循环创建相同长度字符串的数组

【英文标题】：loop to create array of arrays of same-length strings

【发布时间】：2020-02-12 12:09:34

【问题描述】：

我正在尝试创建一个数组数组，这些数组根据它们的长度（字符串）分组，如下面的数组：

var testA = ["carts","carts","cars","cars","ca","ca", "ca","a","a","a","a"];

会变成：

var sortedArrays = [["carts","carts"], ["cars","cars"],["ca","ca","ca"], ["a","a","a","a"]]

我目前正在尝试下面的循环，但根本无法让它工作。我知道这将是一件简单的事情，但我花了很长时间试图完成它，但事实证明这非常令人沮丧。

 var testA = ["carts","carts","cars","cars","ca","ca", "ca","a","a","a","a"];
 var sortedArrays = [];

 for(i=0; i < testA.length; i++)
     longestWord = 0;
     wordHolder = []
     wordHolder.push(testA[i])    

 if (testA[i].length > longestWord)
     longestWord = testA[i].length
     wordHolder.push(testA[i])
 
 sortedArrays.push(wordHolder)

 

任何帮助将不胜感激

【问题讨论】：

javascript？另外，我们得到 testA 已排序？

wordHolder = [] 无法在每次看到一个新词（可能是同一个词）时设置

【参考方案1】：

下面的代码怎么样（Swift，但用其他语言重新编写它应该很容易）：

let testA = ["carts","carts","cars","cars","ca","ca", "ca","a","a","a","a"]
var maxLength = 0
for string in testA 
    if string.count > maxLength  maxLength = string.count 

var sortedArrays: [[String]] = []
for nextLength in 0 ... maxLength 
    let invertedLength = maxLength - nextLength
    let nextArray = testA.filter  $0.count == invertedLength 
    sortedArrays.append(nextArray)

此代码还创建空数组，当然可以轻松跳过。

当然，问题是您的要求是什么。上面的代码很紧凑，但对于大型数组来说速度并不快... 对于大数组，最好先对testA进行排序，然后提取元素。

【参考方案2】：

坚持使用 java 脚本，我能够做到以下几点。我的假设是单词已经按长度降序排序，并且每个长度只有一个单词。那是基于你在做什么。我认为发生的事情是每次循环都被初始化或重置太多。

var testA = ["carts", "carts", "cars", "cars", "ca", "ca", "ca", "a", "a", "a", "a"];
var sortedArrays = [];
var wordHolder = []; // took outside loop to ensure initialized
var curLength = testA[0].length;// assuming here that testA not empty
for (var i = 0; i < testA.length; i++) 
  if (testA[i].length === curLength)  // more of the same
    wordHolder.push(testA[i]);
   else  // new block starts
    curLength = testA[i].length;
    sortedArrays.push(wordHolder);
    // alert (wordHolder)
    // alert (sortedArrays)
    wordHolder = [];
    wordHolder.push(testA[i]);
  

sortedArrays.push(wordHolder);// add the last one

【讨论】：

这成功了！感谢 Jeremy，尤其是注释，它们为我澄清了一些事情。