451. Sort Characters By Frequency

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.

一刷
题解：
O(nlogn), 比较型sort

class Solution {
    class Entry{
        char ch;
        int count;
        
        public Entry(){
            this.count = 0;
        }
    }
    
    public String frequencySort(String s) {
        if(s == null || s.length()<=1) return s;
        Entry[] freq = new Entry[256];
        for(int i=0; i<256; i++){
            freq[i] = new  Entry();
        }
        
        for(int i=0; ib.count-a.count);
        StringBuilder res = new StringBuilder();
        for(int i=0; i0) {
                   freq[i].count--;
                   res.append(freq[i].ch);
               }
            }
        }
        return res.toString();
    }
}

follow up: 如果count一样，按照在字符串中出现顺序输出：解决方案，在entry中添加一个appear(第一次出现时char index)顺序域, sort时，如果count一样，appear小的在前。

class Solution {
    class Entry{
        char ch;
        int count;
        int appear;
        
        public Entry(){
            this.count = 0;
        }
    }
    
    public String frequencySort(String s) {
        if(s == null || s.length()<=1) return s;
        Entry[] freq = new Entry[256];
        for(int i=0; i<256; i++){
            freq[i] = new  Entry();
        }
        
        for(int i=0; ib.count == a.count? (a.appear - b.appear):(b.count - a.count));
        StringBuilder res = new StringBuilder();
        for(int i=0; i0) {
                   freq[i].count--;
                   res.append(freq[i].ch);
               }
            }
        }
        return res.toString();
    }
}

O(n)的方法， bucketsort

public String frequencySort(String s) {
    if (s == null) {
        return null;
    }
//character->count
    Map map = new HashMap();
    char[] charArray = s.toCharArray();
    int max = 0;
    for (Character c : charArray) {
        if (!map.containsKey(c)) {
            map.put(c, 0);
        }
        map.put(c, map.get(c) + 1);
        max = Math.max(max, map.get(c));
    }

    List[] array = buildArray(map, max);

    return buildString(array);
}

private List[] buildArray(Map map, int maxCount) {
    List[] 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[] array) {
    StringBuilder sb = new StringBuilder();
    for (int i = array.length - 1; i > 0; i--) {
        List list = array[i];
        if (list != null) {
            for (Character c : list) {
                for (int j = 0; j < i; j++) {
                    sb.append(c);
                }
            }
        }
    }
    return sb.toString();
}

另一种角度的bucketsort
map>, count->characters

class Solution {
    public String frequencySort(String s) {
        int max = 0;
        char[] arr = s.toCharArray();
        
        int[] count = new int[256];
        for(char c : arr) count[c]++;
        
        Map> map = new HashMap<>();//count->characters
        for(int i=0; i<256; i++){
            if(count[i]!=0){
                int cn = count[i];
                if(!map.containsKey(cn)){
                    map.put(cn, new ArrayList());
                }
                map.get(cn).add((char)i);
                max = Math.max(max, cn);
            }
        }
        
        StringBuilder sb = new StringBuilder();
        for(int cnt = max; cnt>0; cnt--){//the maximum count is 
            if(map.containsKey(cnt)){
                List val = map.get(cnt);
                for(char ch : val){
                    for(int i=0; i