LeetCode-in-Java
451. Sort Characters By Frequency
Medium
Given a string s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string.
Return the sorted string. If there are multiple answers, return any of them.
Example 1:
Input: s = “tree”
Output: “eert”
Explanation: ‘e’ appears twice while ‘r’ and ‘t’ both appear once. So ‘e’ must appear before both ‘r’ and ‘t’. Therefore “eetr” is also a valid answer.
Example 2:
Input: s = “cccaaa”
Output: “aaaccc”
Explanation: Both ‘c’ and ‘a’ appear three times, so both “cccaaa” and “aaaccc” are valid answers. Note that “cacaca” is incorrect, as the same characters must be together.
Example 3:
Input: s = “Aabb”
Output: “bbAa”
Explanation: “bbaA” is also a valid answer, but “Aabb” is incorrect. Note that ‘A’ and ‘a’ are treated as two different characters.
Constraints:
1 <= s.length <= 5 * 105sconsists of uppercase and lowercase English letters and digits.
Solution
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.TreeMap;
public class Solution {
public String frequencySort(String s) {
Map<Character, Integer> map = new HashMap<>();
for (char c : s.toCharArray()) {
map.put(c, map.getOrDefault(c, 0) + 1);
}
TreeMap<Integer, List<Character>> reverseMap = new TreeMap<>(Collections.reverseOrder());
for (Map.Entry<Character, Integer> c : map.entrySet()) {
int freq = map.get(c.getKey());
reverseMap.computeIfAbsent(freq, k -> new ArrayList<>());
reverseMap.get(freq).add(c.getKey());
}
StringBuilder sb = new StringBuilder();
for (Map.Entry<Integer, List<Character>> freq : reverseMap.entrySet()) {
List<Character> list = reverseMap.get(freq.getKey());
for (char c : list) {
for (int i = 0; i < freq.getKey(); i++) {
sb.append(c);
}
}
}
return sb.toString();
}
}