字符串的完美度

题目详情:

我们要给每个字母配一个1-26之间的整数，具体怎么分配由你决定，但不同字母的完美度不同，

而一个字符串的完美度等于它里面所有字母的完美度之和，且不在乎字母大小写，也就是说字母F和f的完美度是一样的。

现在给定一个字符串，输出它的最大可能的完美度。

例如：dad，你可以将26分配给d，25分配给a，这样整个字符串最大可能的完美度为77。

函数头部

int perfect(const char *s);

C++

int perfect(const string &s);

java

public static int perfect(String s);

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace 取反
{
    class Program
    {
        
        static void Main(string[] args)
        {
            string str = "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";
            Console.WriteLine(perfect(str));
            Console.ReadKey();

        }
        static int perfect(string s)
        {
            List
  
    list = new List
   
    (); s= s.ToLower(); Dictionary
    
      dic = new Dictionary
     
      (); for (int i = 0; i < s.Length; i++) { if (!dic.ContainsKey(s[i])) { dic.Add(s[i], 1); } else { dic[s[i]] += 1; } } int res = 0; int num = 26; foreach (var value in dic.Values) { list.Add(value); } int[] nums = list.ToArray(); BubSort(nums); foreach (var item in nums) { res += item * num--; } return res; } /// 
       /// 冒泡排序 ///  /// 
       public static void BubSort(int[] nums) { for (int i = 0; i < nums.Length - 1; i++) { for (int j = 0; j < nums.Length - i - 1; j++) { if (nums[j + 1] > nums[j]) { int temp = 0; temp = nums[j + 1]; nums[j + 1] = nums[j]; nums[j] = temp; } } } } } }