Experience is very important.

  Talk about today do practice it, it's worth mentioning.

  Today do is, without changing the original order, a list deduplication.

  After this idiot thinking and operating three hours to complete the preparation of the following codes:

chong = ["a","a","a","a","a","a","a","a","a","a","a","a","a","a","b","b","b","b","b","b","s","a","a","a","a","a","a","a","a","c","c","c","c","c","c","c","c","c","c","s","b","s"]
chong_T = chong[::-1]
k = 0
a = []
for i in range(0,len(chong)):
    b = chong.count(chong[i])
    a.append(b)
while k <= max(a):
    s = 0
    t = 0
    while s < len(chong_T):
        if chong_T.count(chong_T[s]) > 1:
            chong_T.remove(chong_T[s])
            s += 1
        else:
            s += 1
    while t < len(chong_T):
        if chong_T.count(chong_T[t]) > 1:
            chong_T.remove(chong_T[t])
            t += 1
        else:
            t += 1
    k += 1
chong_T = chong_T[::-1]
print(chong_T)

    Well of course, the first thought is to use a certain count () to count, if there are duplicate, directly remove () away. However, this idea is not perfect, because changing the length of the list, corresponding to a fixed element also changes the index value, so that, by traversing the index value to s, where the jumping occurs. This situation will jump consecutive sufficient number of elements at a same time appears.

    This imperfect method can be performed by "processing" perfect, that is, after repeated sufficiently many times, we can achieve what we want to re-effect, so while the outer loop is selected, the number of cycles of the original list is selected most elements of the "maximum" appears.

    Since then, I asked a friend the great God, people come to me to throw this directly:

    

chong = ["a","a","a","a","a","a","a","a","a","a","a","a","a","a","b","b","b","b","b","b","s","a","a","a","a","a","a","a","a","c","c","c","c","c","c","c","c","c","c","s","b","s"]
for x in chong:
    for y in range(len(chong)-1,chong.index(x),-1):
        if chong[y] == x:
            chong.pop(y)
print(chong)

    Just a few lines to get, do not need to count, just look at the elements are the same on it.

    This idiot can not help but issued a sigh, experienced sophisticate is to force, although both realize the function, but is still as simple as possible.

    To practice more of it.