題目描述
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.
解題思路
非常容易的一道題，trick在于buildArray，找到max count以后建立一個以此為size的array然后把放入對應(yīng)的字母
Java代碼實(shí)現(xiàn)
class Solution {
public String frequencySort(String s) {
if (s == null || s.length() == 0) {
return s;
}
int maxCount = 0;
Map<Character, Integer> charFreqMap = new HashMap<>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
Integer charCount = charFreqMap.get(c);
if (charCount == null) {
charCount = 0;
}
charCount ++;
charFreqMap.put(c, charCount);
maxCount = Math.max(maxCount, charCount);
}
List<Character>[] array = buildArray(charFreqMap, maxCount);
return buildString(array);
}
private List<Character>[] buildArray(Map<Character, Integer> map, int maxCount) {
List<Character>[] array = new List[maxCount + 1];
for (Character c : map.keySet()) {
int count = map.get(c);
if (array[count] == null) {
array[count] = new ArrayList();
}
array[count].add(c);
}
return array;
}
private String buildString(List<Character>[] array) {
StringBuilder sb = new StringBuilder();
for (int i = array.length - 1; i > 0; i--) {
List<Character> list = array[i];
if (list != null) {
for (Character c : list) {
for (int j = 0; j < i; j++) {
sb.append(c);
}
}
}
}
return sb.toString();
}
}
