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leetcode451. Sort Characters By Frequency

feng409 / 2894人閱讀

摘要:題目要求將字符串按照每個字母出現的次數,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。這里要注意大小寫敏感。以此循環,直到將所有的字母都輸出。

題目要求

</>復制代碼

  1. Given a string, sort it in decreasing order based on the frequency of characters.
  2. Example 1:
  3. Input:
  4. "tree"
  5. Output:
  6. "eert"
  7. Explanation:
  8. "e" appears twice while "r" and "t" both appear once.
  9. So "e" must appear before both "r" and "t". Therefore "eetr" is also a valid answer.
  10. Example 2:
  11. Input:
  12. "cccaaa"
  13. Output:
  14. "cccaaa"
  15. Explanation:
  16. Both "c" and "a" appear three times, so "aaaccc" is also a valid answer.
  17. Note that "cacaca" is incorrect, as the same characters must be together.
  18. Example 3:
  19. Input:
  20. "Aabb"
  21. Output:
  22. "bbAa"
  23. Explanation:
  24. "bbaA" is also a valid answer, but "Aabb" is incorrect.
  25. Note that "A" and "a" are treated as two different characters.

將字符串按照每個字母出現的次數,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。這里要注意大小寫敏感。

思路和代碼

直觀的來說,如果可以記錄每個字母出現的次數,再按照字母出現的次數從大到小對字母進行排序,然后順序構成一個新的字符串即可。這里采用流的方式進行排序,代碼如下:

</>復制代碼

  1. public String frequencySort(String s) {
  2. if(s==null || s.isEmpty() || s.length() <= 1) return s;
  3. Map map = new HashMap<>();
  4. for(char c : s.toCharArray()) {
  5. map.put(c, map.getOrDefault(c, new StringBuilder()).append(c));
  6. }
  7. StringBuilder result = map
  8. .values()
  9. .stream()
  10. .sorted((sb1, sb2) -> {
  11. return sb2.length() - sb1.length();
  12. })
  13. .reduce((sb1, sb2) -> sb1.append(sb2))
  14. .get();
  15. return result.toString();
  16. }

如果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,并再次從剩下的字母中選出次數最多的字母。以此循環,直到將所有的字母都輸出。代碼如下:

</>復制代碼

  1. public String frequencySort(String s) {
  2. char[] charArr = new char[128];
  3. for(char c : s.toCharArray())
  4. charArr[c]++;
  5. StringBuilder sb = new StringBuilder();
  6. while(sb.length() < s.length()) {
  7. char maxChar = 0;
  8. for(char charCur = 0; charCur < charArr.length; charCur++) {
  9. if(charArr[charCur] > charArr[maxChar]) {
  10. maxChar = charCur;
  11. }
  12. }
  13. while(charArr[maxChar] > 0){
  14. sb.append(maxChar);
  15. charArr[maxChar]--;
  16. }
  17. }
  18. return sb.toString();
  19. }

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