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 }
原文:https://www.cnblogs.com/EdwardLiu/p/11669473.html