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.
Analysis:
The idea is simple: count the frequency of each character, and then sort the characters by frequency and out put the characters. We can use a two dimension array for the easy counting and sorting.
public String frequencySort(String s) {
int[][] count = new int[256][2];
for(char c: s.toCharArray()) {
count[c][0] = c;
count[c][1]++;
}
Arrays.sort(count, new Comparator<int[]>() {
public int compare(int[] a, int[] b) {
return a[1] == b[1] ? 0 : (a[1] < b[1] ? 1 : -1);
}
});
StringBuilder sb = new StringBuilder();
for(int i=0; i<256; i++) {
if (count[i][1] > 0) {
for(int j=0; j<count[i][1]; j++) {
sb.append((char)count[i][0]);
}
}
}
return sb.toString();
} 