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LeetCode-451-Sort-Characters-By-Frequency.java
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LeetCode-451-Sort-Characters-By-Frequency.java
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class Solution {
// 1. Using Arrays.sort(), time O(NlogN), there is some problem with the sorting algorithms
// public String frequencySort(String s) {
// char[] temp = s.toCharArray();
// Character[] chars = new Character[temp.length];
// for (int i = 0; i < temp.length; i++) {
// chars[i] = temp[i];
// }
// HashMap<Character, Integer> map = new HashMap<>();
// for (char c : chars) {
// map.put(c, map.getOrDefault(c, 0) + 1);
// }
// Arrays.sort(chars, (a, b) -> map.getOrDefault(a, 0) == map.getOrDefault(b, 0) ? Character.compare(a, b) : map.get(b) - map.get(a));
// // Arrays.sort(chars, new Comparator<Character>() {
// // @Override
// // public int compare(Character a, Character b) {
// // if (map.getOrDefault(a, 0) == map.getOrDefault(b, 0)) {
// // return a - b;
// // } else {
// // return map.get(b) - map.get(a);
// // }
// // }
// // });
// StringBuilder sb = new StringBuilder();
// for (Character c : chars) {
// sb.append(c);
// }
// return sb.toString();
// // PriorityQueue<Character> pq = new PriorityQueue<>((a, b) -> map.getOrDefault(a, 0) == map.getOrDefault(b, 0) ? a - b : map.get(b) - map.get(a));
// // for (char c : map.keySet()) {
// // pq.offer(c);
// // }
// // StringBuilder sb = new StringBuilder();
// // while (!pq.isEmpty()) {
// // sb.append(pq.poll());
// // }
// // return sb.toString();
// }
// 2. Using PriorityQueue
/*
Time: O(NlogM) - N is number of chars in the string, M is 26
Runtime: 49 ms, faster than 30.14% of Java online submissions for Sort Characters By Frequency.
Memory Usage: 38.2 MB, less than 96.30% of Java online submissions for Sort Characters By Frequency.
*/
// public String frequencySort(String s) {
// char[] chars = s.toCharArray();
// HashMap<Character, Integer> map = new HashMap<>();
// for (char c : chars) {
// map.put(c, map.getOrDefault(c, 0) + 1);
// }
// PriorityQueue<Character> pq = new PriorityQueue<>((a, b) -> map.getOrDefault(a, 0) == map.getOrDefault(b, 0) ? a - b : map.get(b) - map.get(a));
// for (char c : map.keySet()) {
// pq.offer(c);
// }
// StringBuilder sb = new StringBuilder();
// while (!pq.isEmpty()) {
// char c = pq.poll();
// for (int i = 0; i < map.get(c); i++) {
// sb.append(c);
// }
// }
// return sb.toString();
// }
// 3. Bucket Sort
/*
Time: O(N)
Runtime: 14 ms, faster than 87.32% of Java online submissions for Sort Characters By Frequency.
Memory Usage: 38.7 MB, less than 88.89% of Java online submissions for Sort Characters By Frequency.
*/
public String frequencySort(String s) {
char[] chars = s.toCharArray();
HashMap<Character, Integer> map = new HashMap<>();
for (char c : chars) {
map.put(c, map.getOrDefault(c, 0) + 1);
}
List<Character>[] bucket = new List[s.length() + 1];
for (char c : map.keySet()) {
if (bucket[map.get(c)] == null) {
bucket[map.get(c)] = new ArrayList<>();
}
bucket[map.get(c)].add(c);
}
StringBuilder sb = new StringBuilder();
for (int i = bucket.length - 1; i >= 0; i--) {
if (bucket[i] == null) continue;
for (char c : bucket[i]) {
for (int k = 0; k < i; k++) {
sb.append(c);
}
}
}
return sb.toString();
}
}