Microsoft OA
发布时间:2021-01-08 13:54:50 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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