c++ hash_set 删除原数组重复数据，且不改变原数组顺序

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在中兴笔试中，有一题目要求，去除正整数序列数组中的重复元素，且不改变元素原来位置，例如 arr [] = {1,2,3,4,1,2,3,4} 输出为{1,2,3,4}.再例如arr[] = {1,2,2,2,3,3,4,4,5,5,1,1,1,4,4,4,5,5,6,6}

输出为{1,2,3,4，5,6}

在hash_set中，可以保证元素的唯一性，但是不能够保证元素的顺序，故此题不能采用hash_set解决。

#include <iostream>#include <vector>#include <set>#include <hash_set>using namespace std ;int main(){int arr[] = {1,1,2,2,3,3,4,4,5,5,6,6,1,0,88,99,5,77};int arrLength = sizeof(arr)/sizeof(arr[0]);/*数组元素输出*/cout<<" the arr is : "<< endl;for (int i = 0 ; i < arrLength ; i++){cout<< arr[i]<< " "; }cout<<endl;/*容器输出*/cout<<"the vector test:"<<endl;vector<int> vec(arr,arr+arrLength) ;vector<int>::iterator vecIt ;for (vecIt =vec.begin();vecIt != vec.end();vecIt++){cout<<*vecIt<<" ";}cout<<endl;/*set集合*/set<int> s ;set<int>::iterator setIterator ;for (vecIt = vec.begin() ; vecIt != vec.end();vecIt++){s.insert(*vecIt);}cout<< "set test :"<< endl;for(setIterator = s.begin();setIterator != s.end();setIterator++){cout<<*setIterator<<" ";}cout<<endl;/*hash_set集合元素*/hash_set<int> hs ;hash_set<int>::iterator hash_setInterator ;for (vecIt = vec.begin();vecIt != vec.end();vecIt++){hs.insert(*vecIt);}cout<<"hash set test :"<<endl;for (hash_setInterator = hs.begin() ; hash_setInterator != hs.end();hash_setInterator++){cout<< *hash_setInterator << " ";}system("pause");return 0 ;}

代码输出如下：

要解决中兴的笔试题，可以采用以下思路：

#include <iostream>#include <vector>using namespace  std ;int main(){int arr[] = {1,2,20,1,1,2,2,3,4,5,6,7,8,8,8,9,2,3,4,7,7,8,6,1} ; int arrLength = sizeof(arr)/sizeof(arr[0]);int index = 1 ;bool isExit = false;for(int i = 0 ; i < arrLength ; i++){for (int j = 0 ; j < index ;j++ ){if(arr[i] == arr[j]){isExit = true ;}}if (!isExit){arr[index]= arr[i];index++;}isExit = false ;} for (int i = 0 ;i < index ;i++){cout<<arr[i]<<" ";}    system("pause");return 0 ;}

输出为：

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