leetcode451. Sort Characters By Frequency

題目要求

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.

將字符串按照每一個字母出現的次數，按照出現次數越多的字母組成的子字符串越靠前，生成一個新的字符串。這裏要注意大小寫敏感。

思路和代碼

直觀的來講，若是能夠記錄每一個字母出現的次數，再按照字母出現的次數從大到小對字母進行排序，而後順序構成一個新的字符串便可。這裏採用流的方式進行排序，代碼以下：

public String frequencySort(String s) {
        if(s==null || s.isEmpty() || s.length() <= 1) return s;
        Map<Character, StringBuilder> map = new HashMap<>();
        for(char c : s.toCharArray()) {
            map.put(c, map.getOrDefault(c, new StringBuilder()).append(c));
        }
        StringBuilder result = map
                .values()
                .stream()
                .sorted((sb1, sb2) -> {
                    return sb2.length() - sb1.length();
                })
                .reduce((sb1, sb2) -> sb1.append(sb2))
                .get();
        return result.toString();
    }

若是不直觀的進行排序的話，則每次只要從記錄字母出現次數的map中找出出現次數最多的字母，將其輸出，並再次從剩下的字母中選出次數最多的字母。以此循環，直到將全部的字母都輸出。代碼以下：

public String frequencySort(String s) {
        char[] charArr = new char[128];
        
        
        for(char c :  s.toCharArray()) 
            charArr[c]++;
        
        StringBuilder sb = new StringBuilder();

        
        while(sb.length() < s.length()) {
                    char maxChar = 0;
            for(char charCur = 0; charCur < charArr.length; charCur++) {
                
                if(charArr[charCur] > charArr[maxChar]) {
                    maxChar = charCur;
                }
            }
            while(charArr[maxChar] > 0){
                sb.append(maxChar);
                charArr[maxChar]--;
            }
           
        }
        
        return sb.toString();
    }