删除重复的数组，而preserving顺序C ++ [英] Removing duplicates in an array while preserving the order in C++

问题描述

可能重复：
  <一href="http://stackoverflow.com/questions/1453333/how-to-make-elements-of-vector-unique-remove-non-adjacent-duplicates">How使矢量元素独特之处？ （除去非相邻重复）

有其设置为STL算法，可以从阵列中删除重复项，同时preserving订单部分的任何标准算法。举例来说，如果我有这样一个数组 INT一个[] = {2,1,3,1,4,2}; 去除重复它应该是在 A [] = {2,1,3,4}; 。我不能使用的std ::独特作为数组排序。其他的解决方案，如将其插入的std ::设为我失去的元素将得到排序的顺序。有算法，我可以使用或我要code我自己的任何其他组合？

Is there any standard algorithm which is provided as part of STL algorithms which can remove duplicates from an array while preserving the order. For example, if I have an array like int a[] = {2,1,3,1,4,2}; after the removal of duplicates it should be a[] = {2,1,3,4};. I can not use std::unique as the array is not sorted. Other solutions like inserting it into an std::set I lose the order as the elements will get sorted. Is there any other combination of algorithms I can use or do I have to code my own?

推荐答案

没有标准算法，这一点，但它很容易实现。其原理是保持的std ::设为您已经看到，到目前为止，并跳过重复，同时复制到一个新的向量或数组中的项目。这个工作在为O（n LG电子n）时间及O（n）的内存。如果你使用的C ++ 0x，你可以把它降到O（n）的时间用的std :: unordered_set 的看到，项目设置;这款采用二叉树的哈希表来代替，而应该快了。

There is no standard algorithm for this, but it's fairly easy to implement. The principle is to keep a std::set of the items you've seen so far, and skip duplicates while copying to a new vector or array. This operates in O(n lg n) time and O(n) memory. If you're using C++0x, you can get it down to O(n) time by using std::unordered_set for the seen-items set; this uses a hash table instead of binary trees and should be faster.

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