哈希（5）

/**
* Created by gouthamvidyapradhan on 25/03/2017.
* Given a string, sort it in decreasing order based on the frequency of characters.
* <p>
* Example 1:
* <p>
* Input:
* "tree"
* <p>
* Output:
* "eert"
* <p>
* 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:
* <p>
* Input:
* "cccaaa"
* <p>
* Output:
* "cccaaa"
* <p>
* Explanation:
* Both 'c' and 'a' appear three times, so "aaaccc" is also a valid answer.
* Note that "cacaca" is incorrect, as the same characters must be together.
* <p>
* Example 3:
* <p>
* Input:
* "Aabb"
* <p>
* Output:
* "bbAa"
* <p>
* Explanation:
* "bbaA" is also a valid answer, but "Aabb" is incorrect.
* Note that 'A' and 'a' are treated as two different characters.
*/

public class SortCharByFrequency {
class Freq {
int i;
int c;
}

private int[] buff = new int[256];

/**
* Main method
*
* @param args
* @throws Exception
*/
public static void main(String[] args) throws Exception {
System.out.println(new SortCharByFrequency().frequencySort("askdfkasdkfasdkljfklasdjfkl"));
}

4000
public String frequencySort(String s) {
if (s == null || s.isEmpty()) return s;
Arrays.fill(buff, 0);
StringBuilder sb = new StringBuilder();
for (int i = 0, l = s.length(); i < l; i++)
buff[s.charAt(i)]++;

List<Freq> fList = new ArrayList<>();
for (int i = 0; i < 256; i++) {
if (buff[i] > 0) {
Freq f = new Freq();
f.i = i;
f.c = buff[i];
fList.add(f);
}
}

Collections.sort(fList, (o1, o2) -> Integer.compare(o2.c, o1.c));

for (Freq f : fList) {
char c = (char) f.i;
int freq = f.c;
while (freq-- > 0)
sb.append(c);
}
return sb.toString();
}
}