Microsoft OA
发布时间:2021-01-08 02:58 所属栏目:52 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
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 %d\n",res); 29 int res2 = charCountToDelete("aaaabbbb"); 30 System.out.printf("result is %d\n",res2); 31 int res3 = charCountToDelete("aaaabbbccd"); 32 System.out.printf("result is %d\n",res3); 33 } 34 } (编辑:ASP站长网) |
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