Microsoft OA
发布时间:2021-01-08 06:08:23  所属栏目:系统  来源:网络整理 
            导读:Given a string S consisting of N lowercase letters, return the minimum number of letters that must be deleted to obtain a word in which every letter occurs a unique number of times 没有想到要再存一个HashMap,appearance to character mapping
                
                
                
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 Given a string S consisting of N lowercase letters,return the minimum number of letters that must be deleted to obtain a word in which every letter occurs a unique number of times 没有想到要再存一个HashMap,appearance to character mapping  1 package UniqueCharacter;
 2 
 3 import java.util.HashMap;
 4 import java.util.Map;
 5 
 6 public class Solution {
 7     public static int charCountToDelete(String s) {
 8         HashMap<Character,Integer> map = new HashMap<>();
 9         for (char c : s.toCharArray()) {
10             map.put(c,map.getOrDefault(c,0) + 1);
11         }
12 
13         int res = 0;
14         HashMap<Integer,Character> intToCharMap = new HashMap<>();
15         for (Map.Entry<Character,Integer> entry : map.entrySet()) {
16             int value = entry.getValue();
17             while (intToCharMap.containsKey(value)) {
18                 res ++; // need to delete
19                 value --;
20             }
21             intToCharMap.put(value,entry.getKey());
22         }
23         return res;
24     }
25 
26     public static void main(String[] args) {
27         int res = charCountToDelete("aaaabbbbcccdde");
28         System.out.printf("result is %dn",res);
29         int res2 = charCountToDelete("aaaabbbb");
30         System.out.printf("result is %dn",res2);
31         int res3 = charCountToDelete("aaaabbbccd");
32         System.out.printf("result is %dn",res3);
33     }
34 } 
        
            
        	
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