题目要求
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.将字符串按照每个字母出现的次数，按照出现次数越多的字母组成的子字符串越靠前，生成一个新的字符串。这里要注意大小写敏感。
思路和代码
直观的来说，如果可以记录每个字母出现的次数，再按照字母出现的次数从大到小对字母进行排序，然后顺序构成一个新的字符串即可。这里采用流的方式进行排序，代码如下：
public String frequencySort(String s) {if(s==null || s.isEmpty() || s.length() <= 1) return s;
Map<Character, StringBuilder> map = new HashMap<>();
for(char c : s.toCharArray()) {map.put(c, map.getOrDefault(c, new StringBuilder()).append(c));
}
StringBuilder result = map
.values()
.stream()
.sorted((sb1, sb2) -> {return sb2.length() - sb1.length();})
.reduce((sb1, sb2) -> sb1.append(sb2))
.get();
return result.toString();}如果不直观的进行排序的话，则每次只要从记录字母出现次数的 map 中找出出现次数最多的字母，将其输出，并再次从剩下的字母中选出次数最多的字母。以此循环，直到将所有的字母都输出。代码如下：
public String frequencySort(String s) {char[] charArr = new char[128];
for(char c : s.toCharArray())
charArr++;
StringBuilder sb = new StringBuilder();
while(sb.length() < s.length()) {
char maxChar = 0;
for(char charCur = 0; charCur < charArr.length; charCur++) {if(charArr[charCur] > charArr[maxChar]) {maxChar = charCur;}
}
while(charArr[maxChar] > 0){sb.append(maxChar);
charArr[maxChar]--;
}
}
return sb.toString();}