LeetCode：451

Given a string, sort it in decreasing order based on the frequency of characters.

my answer

class Solution:
    def frequencySort(self, s):
        """
        :type s: str
        :rtype: str
        """
        checked = []
        count = []
        for i in range(len(s)):
            if s[i] not in checked:
                checked.append(s[i])
                count.append(1)
            else:
                count[checked.index(s[i])] += 1
        for j in range(1, len(checked)):  # ins_sort
            while j > 0 and count[j-1] < count[j]:
                count[j-1], count[j] = count[j], count[j-1]
                checked[j-1], checked[j] = checked[j], checked[j-1]
                j -= 1
        result = []
        for k in range(len(checked)):
            result.append(checked[k] * count[k])
        return ''.join(result)   # list-->str

1. ''.join(result):将result列表内的元素以‘’为连接符进行连接，返回字符串

2. 用了插入排序对统计的词频进行排序

other's answer

def frequencySort(self, s):
    result = []
    for letter, times in sorted(Counter(s).iteritems(), key = lambda x: x[1], reverse = True):
        result += [letter] * times
    return ''.join(result)

1. Counter(s)返回一个key为被统计对象，value为对应词频的dict

2. 字典的几种形式:

>>> a = dict(one=1, two=2, three=3)
>>> b = {'one': 1, 'two': 2, 'three': 3}
>>> c = dict(zip(['one', 'two', 'three'], [1, 2, 3]))
>>> d = dict([('two', 2), ('one', 1), ('three', 3)])  # 可以用d[0][0]和d[0][1]来对key和value进行访问
>>> e = dict({'three': 3, 'one': 1, 'two': 2})

3. Counter(s).iteritems()：这是python2的方法，python3中用items()来实现,以上面的d的形式返回key和value