有个很诡异的需求,向高手请教最优解

    clist = ['a','b','b','b','c','c','d']
    def map(clist):
	dic = {}
	zlist = []
	for i in clist:
		dic[i] = clist.count(i)
	for key,value in dic.items():
		if value == 1:
			zlist.append(key)
		else:
			zlist.append(key + '*' + str(value))
	return zlist
    ';'.join(map(clist))

来个python的，哈哈

遍历List,可以用Iterator,可以很方便地遍历，
用.equals进行对象比较就行了，相等则计数

有点像压缩。

Ruby的呼呼： (交互模式, >>代表输入命令，=>代表输出)

>>clist = %w{a b b b c c d}  
=>["a", "b", "b", "b", "c", "c", "d"] 

>>temp = {}
>>clist.each{|i| temp[i].nil? ? temp[i] = 1 : temp[i] +=1} 
>>temp
=>{"a"=>1, "b"=>3, "c"=>2, "d"=>1}  

>>temp.map{|k,v| v == 1 ? k : k + v.to_s}.join ';'   
=>"a;b3;c2;d" 

            
写成一个函数的话：

def count(list)
  temp = {}    
  list.each{|i| temp[i].nil? ? temp[i] = 1 : temp[i] +=1}          
  temp.map{|k,v| v == 1 ? k : k + v.to_s}.join ';'
end 

 

可惜python不支持问号表达式，否则可以写成一行

import re

def maketimes(w, s, u):
    if u:
        return '%s*%d' % (w, len(s)/len(u)+1)
    return w

def parselist(str):
    return ';'.join([maketimes(w,s,u) for w,s,u in re.findall(r'(\w+)((,\1)*)', str)])

if __name__ == '__main__':
    print parselist('a,b,b,b,c,c,d')

另外，如果list里面是object的话，假设是通过其toString来比较是否相同，那么

if __name__ == '__main__':
    l = ['a', 'b', 'b', 'b', 'c', 'c', 'd']
    print parselist(','.join(map(str, l)))

javascript 风格：

var a=["a","b","b","b","c","c","d"];
var b={};
for(var c in a)
{	if(b[a[c]]) b[a[c]]++;else b[a[c]]=1;
}
var e="";
for(var d in b)
{	e+=d+(b[d]>1?("*"+b[d]):"")+";";
}
alert(e);

相当于一个最简单的压缩算法，没什么意思。