Http Sessions在Tomcat中的生命周期(Http Sessions life cycle in Tomcat)

 如果我正确理解您的问题，您希望能够通过“密钥”前缀比较元素，而不是整个字符串内容。 如果是这样，实现自定义相等比较器将允许您轻松利用LINQ集算法。 
 
 这个节目...... 
 
class EqCmp : IEqualityComparer<string> {

    public bool Equals(string x, string y) {
        return GetKey(x).SequenceEqual(GetKey(y));
    }

    public int GetHashCode(string obj) {
        // Using Sum could cause OverflowException.
        return GetKey(obj).Aggregate(0, (sum, subkey) => sum + subkey.GetHashCode());
    }

    static IEnumerable<string> GetKey(string line) {
        // If we just split to 3 strings, the last one could exceed the key, so we split to 4.
        // This is not the most efficient way, but is simple.
        return line.Split(new[] { '*' }, 4).Take(3);
    }

}

class Program {

    static void Main(string[] args) {

        var l1 = new List<string> {
            "index1*index1*index1*some text",
            "index1*index1*index2*some text ** test test test",
            "index1*index2*index1*some text",
            "index1*index2*index2*some text",
            "index2*index1*index1*some text"
        };

        var l2 = new List<string> {
            "index1*index1*index2*some text ** test test test",
            "index2*index1*index1*some text",
            "index2*index1*index2*some text"
        };

        var eq = new EqCmp();

        Console.WriteLine("Elements that are both in l1 and l2:");
        foreach (var line in l1.Intersect(l2, eq))
            Console.WriteLine(line);

        Console.WriteLine("\nElements that are in l1 but not in l2:");
        foreach (var line in l1.Except(l2, eq))
            Console.WriteLine(line);

        // Etc...

    }

}
 
 ...打印以下结果： 
 
Elements that are both in l1 and l2:
index1*index1*index2*some text ** test test test
index2*index1*index1*some text

Elements that are in l1 but not in l2:
index1*index1*index1*some text
index1*index2*index1*some text
index1*index2*index2*some text

If I understand your question correctly, you'd like to be able to compare the elements by their "key" prefix, instead by the whole string content. If so, implementing a custom equality comparer will allow you to easily leverage the LINQ set algorithms.
 
This program...
 
class EqCmp : IEqualityComparer<string> {

    public bool Equals(string x, string y) {
        return GetKey(x).SequenceEqual(GetKey(y));
    }

    public int GetHashCode(string obj) {
        // Using Sum could cause OverflowException.
        return GetKey(obj).Aggregate(0, (sum, subkey) => sum + subkey.GetHashCode());
    }

    static IEnumerable<string> GetKey(string line) {
        // If we just split to 3 strings, the last one could exceed the key, so we split to 4.
        // This is not the most efficient way, but is simple.
        return line.Split(new[] { '*' }, 4).Take(3);
    }

}

class Program {

    static void Main(string[] args) {

        var l1 = new List<string> {
            "index1*index1*index1*some text",
            "index1*index1*index2*some text ** test test test",
            "index1*index2*index1*some text",
            "index1*index2*index2*some text",
            "index2*index1*index1*some text"
        };

        var l2 = new List<string> {
            "index1*index1*index2*some text ** test test test",
            "index2*index1*index1*some text",
            "index2*index1*index2*some text"
        };

        var eq = new EqCmp();

        Console.WriteLine("Elements that are both in l1 and l2:");
        foreach (var line in l1.Intersect(l2, eq))
            Console.WriteLine(line);

        Console.WriteLine("\nElements that are in l1 but not in l2:");
        foreach (var line in l1.Except(l2, eq))
            Console.WriteLine(line);

        // Etc...

    }

}
 
...prints the following result:
 
Elements that are both in l1 and l2:
index1*index1*index2*some text ** test test test
index2*index1*index1*some text

Elements that are in l1 but not in l2:
index1*index1*index1*some text
index1*index2*index1*some text
index1*index2*index2*some text