C#如何比较两个词串并指出哪些部分不同 [英] C# How can I compare two word strings and indicate which parts are different

问题描述

例如，如果我有...

string a = "personil";
string b = "personal";

我想得到...

string c = "person[i]l";

但是它不一定是单个字符.我也可以这样...

However it is not necessarily a single character. I could be like this too...

string a = "disfuncshunal";
string b = "dysfunctional";

对于这种情况，我想得到...

For this case I would want to get...

string c = "d[isfuncshu]nal";

另一个例子是...(注意两个词的长度不同.)

Another example would be... (Notice that the length of both words are different.)

string a = "parralele";
string b = "parallel";

string c = "par[ralele]";

另一个例子是...

string a = "ato";
string b = "auto";

string c = "a[]to";

我该怎么做?

两个字符串的长度可以不同.

The length of the two strings can be different.

添加了其他示例.感谢用户 Nenad 的询问.

Added additional examples. Credit goes to user Nenad for asking.

推荐答案

我今天一定很无聊，但我实际上制作了通过所有 4 个案例的 UnitTest(如果您在此期间没有添加更多).

I must be very bored today, but I actually made UnitTest that pass all 4 cases (if you did not add some more in the meantime).

编辑:添加了 2 个边缘情况并修复它们.

Edit: Added 2 edge cases and fix for them.

Edit2:重复多次的字母(以及这些字母上的错误)

Edit2: letters that repeat multiple times (and error on those letters)

[Test]
[TestCase("parralele", "parallel", "par[ralele]")]
[TestCase("personil", "personal", "person[i]l")]
[TestCase("disfuncshunal", "dysfunctional", "d[isfuncshu]nal")]
[TestCase("ato", "auto", "a[]to")]
[TestCase("inactioned", "inaction", "inaction[ed]")]
[TestCase("refraction", "fraction", "[re]fraction")]
[TestCase("adiction", "ad[]diction", "ad[]iction")]
public void CompareStringsTest(string attempted, string correct, string expectedResult)
{
    int first = -1, last = -1;

    string result = null;
    int shorterLength = (attempted.Length < correct.Length ? attempted.Length : correct.Length);

    // First - [
    for (int i = 0; i < shorterLength; i++)
    {
        if (correct[i] != attempted[i])
        {
            first = i;
            break;
        }
    }

    // Last - ]
    var a = correct.Reverse().ToArray();
    var b = attempted.Reverse().ToArray();
    for (int i = 0; i < shorterLength; i++)
    {
        if (a[i] != b[i])
        {
            last = i;
            break;
        }
    }

    if (first == -1 && last == -1)
        result = attempted;
    else
    {
        var sb = new StringBuilder();
        if (first == -1)
            first = shorterLength;
        if (last == -1)
            last = shorterLength;
        // If same letter repeats multiple times (ex: addition)
        // and error is on that letter, we have to trim trail.
        if (first + last > shorterLength)
            last = shorterLength - first;

        if (first > 0)
            sb.Append(attempted.Substring(0, first));

        sb.Append("[");

        if (last > -1 && last + first < attempted.Length)
            sb.Append(attempted.Substring(first, attempted.Length - last - first));

        sb.Append("]");

        if (last > 0)
            sb.Append(attempted.Substring(attempted.Length - last, last));

        result = sb.ToString();
    }
    Assert.AreEqual(expectedResult, result);
}

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