Count Number Pairs

题目要求：
提供一个一维整数数组，数组内的元素取值范围为[0,520]，统计输出数组内的任意两元素之和为520的个数 :::danger 限制：数组内的元素只能使用一次 :::

using System;
using System.Collections.Generic;
namespace CountNumberPairs
{
    internal static class Program
    {
        static void Main(string[] args)
        {
            int[] nums = { 250,520,170,420,100,100,260 };
            int count = Count(nums);
            Console.WriteLine(count);
        }
        /*创建一个字典（初始为空），将数组中的元素与总额的差值数与字典键值对的键值进行比较；
         * 存在该键值，则数量-1，计数+1；且若剩余匹配次数为0时，移除该值；
         * 不存在则将该值导入字典；
         */
        static int Count(int[] nums)
        {
            // 保护逻辑，防止传入数组为空
            if (nums == null)
            {
                return 0;
            }
            int counter = 0;
            var dict = new Dictionary<int, int>();
            foreach (var n in nums)
            {
                int diff = 520 - n;
                if (dict.ContainsKey(diff))
                {
                    counter++;
                    /*以下两行
                     * if（--dict[diff]==0）
                     */
                    dict[diff]--;
                    if (dict[diff] == 0)
                    {
                        dict.Remove(diff);
                    }
                }
                else
                {
                    if (!dict.ContainsKey(n))
                    {
                        dict[n] = 0;
                    }
                    dict[n]++;
                 }
            }
            return counter;
        }
    }
}

using System;
using System.Collections.Generic;
namespace CountNumberPairs
{
    internal static class Program
    {
        static void Main(string[] args)
        {
            int[] nums = new int[10] { 1, 9, 2, 8, 7, 3, 4, 6, 5, 5 };
            int count = Count(nums,10,10);
            Console.WriteLine(count);
        }
        static int Count(int[] nums,int k,int sum)
        {
            if (nums == null)
            {
                return 0;
            }
            int counter = 0;
            var dict = new int[k+1];
            foreach (var n in nums)
            {
                int diff = sum - n;
                if (dict[diff]>0)
                {
                    counter++;
                    dict[diff]--;
                }
                else
                {                    
                    dict[n]++;
                 }                
            }
            return counter;
        }
    }
}