451. Sort Characters By Frequency

题目：

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.

代码：
别看题目很长，其实就是： 给定一个字符串，找出里面每个字符出现的频率，按频率从从大到小排序。
想了半天，觉得怎么都是遍历一遍，记录并求出每个字符出现的次数，之后排序。
可是，网上查了一下，才知道python原来是那么的牛B，有这样一个函数：
看举例子就明白了，而且顺序都排好了，是不是很流弊！

这个函数在collection模块的Counter类中：

于是，看似很复杂的题目，用python写只剩一句话了：

class Solution(object):
    def frequencySort(self, s):
        """
        :type s: str
        :rtype: str
        """
        #print (collections.Counter(s).most_common())
        return ''.join(c * num for c, num in collections.Counter(s).most_common())

当然，我总觉得这样自身没什么思考。不过有时候快速开发，解决问题，确实更重要！