[LeetCode] Sort Characters By Frequency(Java)

我的思想是list数组按顺序构造，讨论中高票使用数组实现，效果更好

我的代码

public class Solution {
public String frequencySort(String s) {
int length = s.length();
List<String>[] sortList = new LinkedList[length+1];
Map<String,Integer> hash = new LinkedHashMap<String,Integer>();
StringBuffer result = new StringBuffer();
for(int i = 0;i < length;i++){
String key = "" + s.charAt(i);
hash.put(key,hash.getOrDefault(key, 0) + 1);
}
for(Map.Entry<String, Integer> entry : hash.entrySet()){
int value = entry.getValue();
if (sortList[value] == null) {
sortList[value] = new LinkedList<String>();
}
sortList[value].add(entry.getKey());
}
for(int i = length;i > 0;i--){
if(sortList[i] == null)
continue;
int len = sortList[i].size();
for(int j = 0; j < len;j++){
String val = sortList[i].get(j);
//System.out.println(val);
for(int k = 0; k < i;k++){
result.append(val);

4000
}
}
}
//System.out.println(result);
return result.toString();
}
}这个是高票代码
public String frequencySort(String s) {
if(s.length() < 3)
return s;
int max = 0;
int[] map = new int[256];
for(char ch : s.toCharArray()) {
map[ch]++;
max = Math.max(max,map[ch]);
}
String[] buckets = new String[max + 1]; // create max buckets
for(int i = 0 ; i < 256; i++) { // join chars in the same bucket
String str = buckets[map[i]];
if(map[i] > 0)
buckets[map[i]] = (str == null) ? "" + (char)i : (str + (char) i);
}
StringBuilder strb = new StringBuilder();
for(int i = max; i >= 0; i--) { // create string for each bucket.
if(buckets[i] != null)
for(char ch : buckets[i].toCharArray())
for(int j = 0; j < i; j++)
strb.append(ch);
}
return strb.toString();
}2016/12/26