题目：

You are given an input string.

For each symbol in the string if it's the first character occurence, replace it with a '1', else replace it with the amount of times you've already seen it...

But will your code be performant enough?

---

Examples:

input   =  "Hello, World!"result  =  "1112111121311"input   =  "aaaaaaaaaaaa"result  =  "123456789101112"

题目大意：将字符出现的次数打印出来，以数字1,2,3，，，表示

 

解题办法：（再想想，也许就出来了）

def numericals(s):    # code    dic = {}    L = []    for i in s:        if i not in dic.keys():            dic[i] = 1            L.append(dic[i])        else:            dic[i] += 1            L.append(dic[i])    return ''.join([str(x) for x in L])

用时：1小时3分51秒，，，真长。。。

看下别人的解法：

1、使用defaultdict

from collections import defaultdictdef numericals(s):    d, lst = defaultdict(int), []    for c in s:        d[c] += 1        lst.append(d[c])    return ''.join(map(str, lst))

2、使用生成器

from collections import Counterdef numericals(s):    def f(s):        cnts = Counter()        for c in s:            cnts[c] += 1            yield cnts[c]    return ''.join(map(str, f(s)))

 

知识点：

1、使用计数count

2、字典可以使用特殊字符作为key

3、将一个list转化为str：

　　3.1、如果list中有数值，需要转为str，可以使用遍历：

str(x) for x in L

　　3.2、可以使用map

map(str, L)

''.join(str(x) for x in L)''.join(map(str, L))