检查两个字符串是否包含相同单词但顺序不同

debugcn 发表于 Dev

克里克·詹姆斯

我需要一种方法来检查两个字符串，以查看它们中是否有相同的单词但位置不同。我只需要单词中没有单词的数量，而不是实际单词。例如：字符串1为“这是乱序测试”，字符串2为“这是乱序测试”将返回1，因为一个单词乱序。

我从以下代码开始：

public int OutOfOrder(string string1, string string2)
{
    var search1 = string1.Split(' ');
    var search2 = string2.Split(' ');
    var OutOfOrder = 0;

    for (int i = 0; i < search1.Count() - 1; i++)
    {
        if (search2.Contains(search1[i]))
        {
            for (int j=0; j < search2.Count()-1; j++)
            {
                if(search1[i] == search2[j] && i == j)
                {
                    continue;
                }
                else if (search1[i] == search2[j] && i != j)
                {
                    OutOfOrder++;
                    break;
                }
            }
        }
    }

    return OutOfOrder;
}

但是，遇到第一个后，一切都乱了，返回的数字不正确。

梅雷布·海尔

做到了

public static int OutOfOrder(string a, string b){

            //not out of order issuex`
        if (a.Length != b.Length) return -1;

        foreach(string str in a.Split(' ')){
            //not out of order issue
            if (!b.Contains(str)) return -1;
        }

        //if out of order issue...

        int count=0;

        string[] awords = a.Split(' ');
        string[] bwords = b.Split(' ');

        for(int i=0; i<bwords.Length; i++){
            if (!awords[i].Equals(bwords[i])) count++;
        }

        return (int)Math.Ceiling(0.0f + count/2);
        //if one is out of order, then it's expected it renders another out of order too.
        //so we need to divide it by 2 to get the one disturbing.
        //and ceiling is used to help for odd number of words
    }

现在测试

public static void Main(string[] args)
{
    string a = "This solves the problem";
    string b = "This solves problem the";

    Console.WriteLine(OutOfOrder(a, b));
}

输出1

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