Leetcode 451. Sort Characters By Frequency 按频率对字符排序 解题报告

1 解题思想

2 原题

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.

3 AC解

public class Solution {
public String frequencySort(String s) {
int frequency[] = new int[256];
for(char c:s.toCharArray()){
frequency[c] ++;
}
// 统计频率，使用Treemap排序,这里白板写的代码，直接将频率取反，这样方便
TreeMap<Integer,String> map= new TreeMap<Integer,String>();
for(int i=0;i<256;i++){
if(frequency[i] == 0) continue;
StringBuilder sb = new StringBuilder(map.getOrDefault(-frequency[i],""));
for(int j=0;j<frequency[i];j++)
sb.append((char)i);
map.put(-frequency[i],sb.toString());
}
StringBuilder sb = new StringBuilder();
Iterator it = map.keySet().iterator();
while(it.hasNext()){
Integer key = (Integer)it.next();
sb.append(map.get(key));
}
return sb.toString();
}
}