题目描述：

LeetCode 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.

题目大意：

给定一个字符串，将字符按照出现次数倒序排列。

解题思路：

字符统计 + 排序

Python代码：

class Solution(object):
    def frequencySort(self, s):
        """
        :type s: str
        :rtype: str
        """
        return ''.join(c * t for c, t in collections.Counter(s).most_common())

Java代码：

public class Solution {
    public String frequencySort(String s) {
        HashMap<Character, Integer> charFreqMap = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            charFreqMap.put(c, charFreqMap.getOrDefault(c, 0) + 1);
        }
        ArrayList<Map.Entry<Character, Integer>> list = new ArrayList<>(charFreqMap.entrySet());
        list.sort(new Comparator<Map.Entry<Character, Integer>>(){
            public int compare(Map.Entry<Character, Integer> o1, Map.Entry<Character, Integer> o2) {
                return o2.getValue().compareTo(o1.getValue());
            }
        });
        StringBuffer sb = new StringBuffer();
        for (Map.Entry<Character, Integer> e : list) {
            for (int i = 0; i < e.getValue(); i++) {
                sb.append(e.getKey());
            }
        }
        return sb.toString();
    }
}

 

本文链接：http://bookshadow.com/weblog/2016/11/02/leetcode-sort-characters-by-frequency/
请尊重作者的劳动成果，转载请注明出处！书影博客保留对文章的所有权利。

如果您喜欢这篇博文，欢迎您捐赠书影博客： ，查看支付宝二维码