python代码(2)---google中国编程挑战赛入围赛真题HardDuplicateRemover(1000分)

问题分析(问题描述见后面):

首先建立一个dup表,来记录duplicate item下一次出现的下标,如果没有下一相同的item,则记为0.

循环dup表:

       当前位置i

       如果没有duplicate值,直接写到输出tlist中.

      有duplicate值:

          寻找 head , tail下标值. head=i(当前循环到的位置), tail=min(head后的每一个没有duplicate 的item的下标,最后duplicate项的下标)

          在 head---tail 这段区域中判断是否有比当前item更小的值:

                是----舍弃当前值

                否----输出到tlist中,将duplicate值标记为已经舍弃.

代码:

def  findtail(dup,i):
     """  find the tail of dupicate items' index """
     if  dup[i] > 0:
         return  findtail(dup,dup[i])
     else :
         return  i
     pass
def  markRemove(dup,i):
     """  i for reserved,then others mark remove(-1) """
    k = dup[i]
    dup[i] =- 1
     if  k > 0:
        markRemove(dup,k)
     pass
def  tmin(list,dup):
     """  find the min in list,exclude the mark remove items in the dup """
    m = 65535
     for  i  in  range(len(dup)):
         if  dup[i] >= 0  and  m > list[i]:
            m = list[i]
     return  m
     pass
def  hdr(list):
     # First create the whether duplicate list,0 for single,sub for multi
    dup = [0] * len(list)
    iterate = range(len(list))
     for  i  in  iterate:
         if  list[i]  in  list[i + 1 :]:
            dup[i] = list[i + 1 :].index(list[i]) + i + 1
             pass
         pass
     # finally Remove the duplicate items
    tlist = []
     for  i  in  iterate:
         if  dup[i] > 0:
            tail = findtail(dup,i)
             if  0  in  dup[i + 1 :tail]:
                tail = dup[i + 1 :].index(0) + i + 1
             if  list[i] <= tmin(list[i + 1 :tail + 1 ],dup[i + 1 :tail + 1 ]):
                tlist.append(list[i])
                markRemove(dup,i)                          
             pass
         elif  dup[i] == 0:
            tlist.append(list[i])
     return  tlist

if   __name__ == " __main__ " :
    list = [ 1 , 2 , 1 , 4555 , 9 , 2 , 78 , 1 , 90 , 111 ]
     print  list
     print   " === " ,hdr(list)
    list = [ 5 , 4 , 6 , 5 , 45 , 4 , 9 , 6 , 78 , 1 , 9 , 45 , 6 , 29 ]
     print  list
     print   " === " ,hdr(list)    

问题描述:
We have a sequence of integers, and we would like to remove all duplicate elements from this sequence. There may be multiple ways to perform this task. For example, given the sequence { 1, 2, 1, 3 }, we could end up with either { 1, 2, 3 } or { 2, 1, 3 } as the remaining sequence, depending on which duplicate 1 we remove from the original sequence. For this problem, we want to return the lexicographically first of of all possible remaining sequences. A sequence S1 comes before sequence S2 lexicographically if and only if S1 has a smaller value than S2 at the lowest index at which the two sequences differ (so, for example, { 1, 2, 3 } comes before { 2, 3, 1 }).
You will be given a int[] sequence. Return a int[] containing the sequence after all the duplicates are removed. See the examples for further clarification.
Definition
Class:
HardDuplicateRemover
Method:
process
Parameters:
int[]
Returns:
int[]
Method signature:
int[] process(int[] sequence)
(be sure your method is public)
Problem Statement

Constraints
-
sequence will have between 1 and 50 elements, inclusive.
-
Each element of sequence will be between 1 and 1000, inclusive.
Examples
0)

{5, 6, 5, 1, 6, 5}
Returns: {1, 6, 5 }
There are six different ways to remove duplicates (remaining numbers are marked by '*'):
{ *5, *6,  5, *1,  6,  5},
{ *5,  6,  5, *1, *6,  5},
{  5, *6, *5, *1,  6,  5},
{  5,  6, *5, *1, *6,  5},
{  5, *6,  5, *1,  6, *5},
{  5,  6,  5, *1, *6, *5}.

The last variant is the lexicographically first.
1)

{3, 2, 4, 2, 4, 4}
Returns: {3, 2, 4 }

2)

{6, 6, 6, 6, 6, 6}
Returns: {6 }

3)

{1, 3, 2, 4, 2, 3}
Returns: {1, 2, 4, 3 }

4)

{5, 4, 1, 5}
Returns: {4, 1, 5 }

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