按正确顺序在字符串数组中查找常见字符 [英] Find common chars in array of strings, in the right order

问题描述

我花了几天时间研究一个函数，以正确的顺序获取字符串数组中的常见字符，以创建通配符.

I spent days working on a function to get common chars in an array of strings, in the right order, to create a wildcard.

这是一个解释我的问题的例子.我做了大约3个函数，但是每次字母绝对位置不同时总是有一个bug.

Here is an example to explain my problem. I made about 3 functions, but I always have a bug when the absolute position of each letter is different.

假设+"是通配符":

Array(
0 => '48ca135e0$5',
1 => 'b8ca136a0$5',
2 => 'c48ca13730$5',
3 => '48ca137a0$5');

应该返回:

$wildcard='+8ca13+0$5';

在这个例子中，棘手的是 $array[2] 比其他字符多 1 个字符.

In this example, the tricky thing is that $array[2] as 1 char more than others.

其他例子:

Array(
0 => "case1b25.occHH&FmM",
1 => "case11b25.occHH&FmM",
2 => "case12b25.occHH&FmM",
3 => "case20b25.occHH&FmM1");

应该返回:

$wildcard='case+b25.occHH&FmM+';

在这个例子中，棘手的部分是:
- 重复字符，例如删除"部分中的 1 -> 11，以及公共部分中的 c -> cc
- $array[2] & 中的2"字符[3]在删除"部分不在同一个位置
- 最后一个字符串末尾的1"字符

In this example, the tricky parts are :
- Repeating chars, such as 1 -> 11 in the "to delete" part, and c -> cc in the common part
- The "2" char in $array[2] & [3] in the "to delete" part is not in the same position
- The "1" char at the end of the last string

我真的需要帮助，因为我找不到此功能的解决方案，而且它是我应用程序的主要部分.

I really need help because I can't find a solution to this function and it is a main part of my application.

提前谢谢，不要犹豫，问问题，我会尽快回答.

Thanks in advance, don't hesitate to ask questions, I will answer as fast as possible.

Mykeul

推荐答案

主要代码:
步骤 1:将字符串按长度从最短到最长排序到数组[]
第 2 步:比较 array[0] 和 array[1] 中的字符串以获得 $temp_wildcard
第 3 步:将数组 [2] 中的字符串与 $temp_wildcard 进行比较以创建新的 $temp_wildcard
第 4 步:继续将每个字符串与 $temp_wildcard 进行比较 - 最后一个 $wildcard 是您的 $temp_wildcard

Main code:
Step 1: Sort strings by length, shortest to longest, into array[]
Step 2: Compare string in array[0] and array[1] to get $temp_wildcard
Step 3: Compare string in array[2] with $temp_wildcard to create new $temp_wildcard
Step 4: Continue comparing each string with $temp_wildcard - the last $wildcard is your $temp_wildcard

好的，现在我们来解决如何比较两个字符串以返回通配符字符串的问题.

OK, so now we're down to the problem of how to compare two strings to return your wildcard string.

子程序代码:逐个字符比较字符串，当比较不匹配时将通配符替换为返回值.

Subroutine code: Compare strings character-by-character, substituting wildcards into your return value when the comparison doesn't match.

为了处理不同长度的问题，对于第二个字符串较长并带有偏移量的每个字符，运行这个比较额外的时间.(比较 string1[x] 和 string2[x+offset].)对于每个返回的字符串，计算通配符的数量.子例程应该返回具有最少通配符字符数的答案.

To handle the problem of different lengths, run this comparison an extra time for each character that the second string is longer with an offset. (Compare string1[x] to string2[x+offset].) For each returned string, count the number of wildcard characters. The subroutine should return the answer with the fewest number of wildcard characters.

祝你好运！

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