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摘要：題目要求將字符串按照每個(gè)字母出現(xiàn)的次數(shù)，按照出現(xiàn)次數(shù)越多的字母組成的子字符串越靠前，生成一個(gè)新的字符串。這里要注意大小寫敏感。以此循環(huán)，直到將所有的字母都輸出。
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.
將字符串按照每個(gè)字母出現(xiàn)的次數(shù)，按照出現(xiàn)次數(shù)越多的字母組成的子字符串越靠前，生成一個(gè)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(gè)字母出現(xiàn)的次數(shù)，再按照字母出現(xiàn)的次數(shù)從大到小對字母進(jìn)行排序，然后順序構(gòu)成一個(gè)新的字符串即可。這里采用流的方式進(jìn)行排序，代碼如下：
public String frequencySort(String s) {
if(s==null || s.isEmpty() || s.length() <= 1) return s;
Map 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();
}
如果不直觀的進(jìn)行排序的話，則每次只要從記錄字母出現(xiàn)次數(shù)的map中找出出現(xiàn)次數(shù)最多的字母，將其輸出，并再次從剩下的字母中選出次數(shù)最多的字母。以此循環(huán)，直到將所有的字母都輸出。代碼如下：
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();
}
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