451. Sort Characters By Frequency

Given a string, sort it in decreasing order based on the frequency of characters.

Example 1:

Input:
"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:
"cccaaa"

Output:
"cccaaa"

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.

Example 3:

Input:
"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.

用Bucket Sort解决问题。思路如下：
1、创建hashmap存相应的字符，

2、创建bucket在相应的框里存字符，

3、后向前遍历添加字符串。

代码如下：

public class Solution {
public String frequencySort(String s) {
HashMap<Character, StringBuilder> hm = new HashMap<Character, StringBuilder>();
char[] chs = s.toCharArray();
int len = chs.length;
StringBuilder[] strs = new StringBuilder[len + 1];
StringBuilder res = new StringBuilder();
for (char ch: chs) {
if (!hm.containsKey(ch)) {
StringBuilder temp = new StringBuilder();
hm.put(ch, temp);
}
hm.get(ch).append(ch);
}
for (char ch: hm.keySet()) {
int lens = hm.get(ch).length();
if (strs[lens] == null) {
strs[lens] = new StringBuilder();
}
strs[lens].append(hm.get(ch));
}
for (int i = len; i >= 0; i--) {
if (strs[i] != null) {
res.append(strs[i]);
}
}
return res.toString();
}
}