字典可能是最接近的。System.Collections.Generic.Dictionary实现了System.Collections.Generic.IDictionary接口(类似于Java的Map接口)。

你应该注意的一些显著差异:

Adding/Getting items Java's HashMap has the put and get methods for setting/getting items myMap.put(key, value) MyObject value = myMap.get(key) C#'s Dictionary uses [] indexing for setting/getting items myDictionary[key] = value MyObject value = myDictionary[key] null keys Java's HashMap allows null keys .NET's Dictionary throws an ArgumentNullException if you try to add a null key Adding a duplicate key Java's HashMap will replace the existing value with the new one. .NET's Dictionary will replace the existing value with the new one if you use [] indexing. If you use the Add method, it will instead throw an ArgumentException. Attempting to get a non-existent key Java's HashMap will return null. .NET's Dictionary will throw a KeyNotFoundException. You can use the TryGetValue method instead of the [] indexing to avoid this: MyObject value = null; if (!myDictionary.TryGetValue(key, out value)) { /* key doesn't exist */ }

Dictionary's有一个ContainsKey方法可以帮助处理前两个问题。

请查看MSDN上关于Hashtable类的文档。

表示基于键的哈希代码组织的键和值对的集合。

另外，请记住，这不是线程安全的。

答案是

字典

看看我的函数，它的简单添加使用了字典中最重要的成员函数

如果列表中包含duplicate项，则此函数返回false

 public static bool HasDuplicates<T>(IList<T> items)
    {
        Dictionary<T, bool> mp = new Dictionary<T, bool>();
        for (int i = 0; i < items.Count; i++)
        {
            if (mp.ContainsKey(items[i]))
            {
                return true; // has duplicates
            }
            mp.Add(items[i], true);
        }
        return false; // no duplicates
    }

我只是想说说我的意见。 这是@Powerlord的回答。

输入“null”而不是空字符串。

private static Dictionary<string, string> map = new Dictionary<string, string>();

public static void put(string key, string value)
{
    if (value == null) value = "null";
    map[key] = value;
}

public static string get(string key, string defaultValue)
{
    try
    {
        return map[key];
    }
    catch (KeyNotFoundException e)
    {
        return defaultValue;
    }
}

public static string get(string key)
{
    return get(key, "null");
}

字典可能是最接近的。System.Collections.Generic.Dictionary实现了System.Collections.Generic.IDictionary接口(类似于Java的Map接口)。

你应该注意的一些显著差异:

Adding/Getting items Java's HashMap has the put and get methods for setting/getting items myMap.put(key, value) MyObject value = myMap.get(key) C#'s Dictionary uses [] indexing for setting/getting items myDictionary[key] = value MyObject value = myDictionary[key] null keys Java's HashMap allows null keys .NET's Dictionary throws an ArgumentNullException if you try to add a null key Adding a duplicate key Java's HashMap will replace the existing value with the new one. .NET's Dictionary will replace the existing value with the new one if you use [] indexing. If you use the Add method, it will instead throw an ArgumentException. Attempting to get a non-existent key Java's HashMap will return null. .NET's Dictionary will throw a KeyNotFoundException. You can use the TryGetValue method instead of the [] indexing to avoid this: MyObject value = null; if (!myDictionary.TryGetValue(key, out value)) { /* key doesn't exist */ }

Dictionary's有一个ContainsKey方法可以帮助处理前两个问题。

让我用一个"codaddict's algorithm"的例子来帮助你们理解

“c#中的字典”在平行宇宙中是“Java中的Hashmap”。

有些实现是不同的。请参阅下面的示例以更好地理解。

声明Java HashMap:

Map<Integer, Integer> pairs = new HashMap<Integer, Integer>();

声明c#字典:

Dictionary<int, int> Pairs = new Dictionary<int, int>();

从位置获取值:

pairs.get(input[i]); // in Java
Pairs[input[i]];     // in C#

在位置处设置值:

pairs.put(k - input[i], input[i]); // in Java
Pairs[k - input[i]] = input[i];    // in C#

从Codaddict的算法下面可以观察到一个整体示例。

Java中的codaddict算法:

import java.util.HashMap;

public class ArrayPairSum {

    public static void printSumPairs(int[] input, int k)
    {
        Map<Integer, Integer> pairs = new HashMap<Integer, Integer>();

        for (int i = 0; i < input.length; i++)
        {
            if (pairs.containsKey(input[i]))
                System.out.println(input[i] + ", " + pairs.get(input[i]));
            else
                pairs.put(k - input[i], input[i]);
        }

    }

    public static void main(String[] args)
    {
        int[] a = { 2, 45, 7, 3, 5, 1, 8, 9 };
        printSumPairs(a, 10);

    }
}

Codaddict的c#算法

using System;
using System.Collections.Generic;

class Program
{
    static void checkPairs(int[] input, int k)
    {
        Dictionary<int, int> Pairs = new Dictionary<int, int>();

        for (int i = 0; i < input.Length; i++)
        {
            if (Pairs.ContainsKey(input[i]))
            {
                Console.WriteLine(input[i] + ", " + Pairs[input[i]]);
            }
            else
            {
                Pairs[k - input[i]] = input[i];
            }
        }
    }
    static void Main(string[] args)
    {
        int[] a = { 2, 45, 7, 3, 5, 1, 8, 9 };
        //method : codaddict's algorithm : O(n)
        checkPairs(a, 10);
        Console.Read();
    }
}