题目地址：https://leetcode.com/problems/sort-characters-by-frequency/description/
题目描述
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
很喜欢这种题目很短，而栗子很多的题目~
题意就是把字符串按照字符出现的次数重新排列。
解题方法
字典
用python真的超级简单呀使用Counter类就能统计每个字符出现的次数，使用most_common函数就能按次序排列，最后字符与其出现的次数相乘就得到了字符串
下面是使用的一个例子的结果：
count = collections.Counter('abbdfas').most_common()
print count
# 输出
[('a', 2), ('b', 2), ('s', 1), ('d', 1), ('f', 1)]
代码：
class Solution(object):
def frequencySort(self, s):
"""
:type s: str
:rtype: str
"""
count = collections.Counter(s).most_common()
res = ''
for c, v in count:
res += c * v
return res
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优先级队列
C++默认的priority_queue是大顶堆。
C++构造把字符重复多次的新字符串的一个方法是append方法，第一个参数是整形表示重复次数，第二个参数是字符。
C++代码如下：
class Solution {
public:
string frequencySort(string s) {
string res;
unordered_map<char, int> count;
for (char c : s) count[c]++;
priority_queue<pair<int, char>> q;
for (auto a : count) q.push({a.second, a.first});
while (!q.empty()) {
auto t = q.top(); q.pop();
res.append(t.first, t.second);
}
return res;
}
};
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排序
参考Grandyang的解法：http://www.cnblogs.com/grandyang/p/6231504.html
我们也可以使用STL自带的sort来做，关键就在于重写comparator，由于需要使用外部变量，记得中括号中放入＆，然后我们将频率大的返回，注意一定还要处理频率相等的情况，要不然两个频率相等的字符可能穿插着出现在结果res中，这样是不对的。参见代码如下。
class Solution {
public:
string frequencySort(string s) {
unordered_map<char, int> m;
for (char c : s) ++m[c];
sort(s.begin(), s.end(), [&](char& a, char& b){
return m[a] > m[b] || (m[a] == m[b] && a < b);
});
return s;
}
};
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