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.
思路：
hashmap求frequency加排序，排序可以用bucket sort or heap sort。
class Solution {
public String frequencySort(String s) {
Map<Character, Integer> map = new HashMap();
int max = 0;
for(char c : s.toCharArray()) {
map.put(c, map.getOrDefault(c, 0) + 1);
max = Math.max(map.get(c), max);
}
List<Character>[] bucket = new List[max + 1];
int count = 0;
for(char c : map.keySet()) {
int index = map.get(c);
if(bucket[index] == null)
bucket[index] = new ArrayList();
bucket[index].add(c);
count++;
}
StringBuilder sb = new StringBuilder();
sb.append("");
for(int i = bucket.length-1; i >= 0; i--) {
if(bucket[i] != null) {
for(char c : bucket[i]) {
for(int j = 0; j < i; j++)
sb.append(c);
count--;
}
if(count == 0)
break;
}
}
return sb.toString();
}
}/**
* @param {string} s
* @return {string}
*/
var frequencySort = function(s) {
const len = s.length
if (!len) {
return s
}
const map = new Map()
const temp = []
let str = ''
for (let i = 0; i < len; i++) {
const char = s[i]
if (!map.has(char)) {
map.set(char, 1)
} else {
map.set(char, map.get(char) + 1)
}
}
map.forEach((val, key) => {
temp.push({ val, key })
})
temp.sort((a, b) => b.val - a.val)
temp.forEach(item => {
str += item.key.repeat(item.val)
})
return str
};
