removeduplicatearray

# remove duplicates from sorted array
def removeduplicate(intls):  #O(n)  空间复杂度略高
    length=len(intls)
    s=[]
    for i in range(0,length-1):
        if intls[i+1]==intls[i]:
            s.append(i+1)
    c=-1
    for x in s:
        c+=1
        intls.pop(x-c)
    return intls
print(removeduplicate([1,1,2,2,6,7]))

def removeduplicate2(intls):
    return list(set(intls))
print(removeduplicate2([1,1,2,2,6,7]))

def removeduplicate3(intls):  #O(n)
    for i in intls:
        if i==intls[intls.index(i)+1]:
            intls.pop(intls.index(i)+1)
    return intls
print(removeduplicate3([1,1,2,3,3,5,5,5,7]))

#Follow up for "Remove Duplicates": What if duplicates are allowed at most twice?
# For example, given sorted array A = [1,1,1,2,2,3] , your function should
# return length = 5 , and A is now [1,1,2,2,3]
#python里的字典就像java里的HashMap，以键值对的方式存在并操作，其特点如下
# 通过键来存取，而非偏移量；	键值对是无序的；	键和值可以是任意对象；

def reomveDup(inls,times):
    hashtable={}
    newls=[]
    for i in set(inls):
        hashtable[i]=0
    for j in inls:
        if hashtable[j]>=times:
            pass
        else:
            hashtable[j]+=1
            newls.append(j)
    return newls
print(reomveDup([1,1,2,2,3,3,3,3,5],2))
#思考为什么需要一个新的列表存储结果而不是直接删除原列表元素
# ls = [1, 2, 3, 4, 5]
# for i in ls:
#     if i == 2:
#         ls.remove(i)
#     else:
#         print(i)
# 
# 1
# 4
# 5   （输出没有3）
# ls
# [1, 3, 4, 5]  （但是里面有3）