如何排序一个数组，得到新数组,并得到新数组每个元素在原数组的下标？

[ljm870605]

假如一个数组a[6]={1,1,4,2,3,2,1};用一种排序方法（最好用堆排序或归并排序等复杂度低一些的方法，因为实际的数组a很大）得到一个新的排序后的数组b[6]={1,1,1,2,2,3,4};有什么办法知道b数组中的元素在a数组中下标的位置?

[oniisama]

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main()
{
	int a[]={1,1,4,2,3,2,1};
	struct ArrayIndex
	{
		int value;
		int index;
	}element;
	vector<ArrayIndex> myarray;
	for(int i=0;i<sizeof(a)/sizeof(int);++i)
	{
		element.value=a[i];
		element.index=i;
		myarray.push_back(element);
	}
	sort(myarray.begin(),myarray.end(),[&](ArrayIndex x,ArrayIndex y){return x.value<y.value;});
	for_each(myarray.begin(),myarray.end(),[&](ArrayIndex x){wcout<<x.index<<endl;});
}

[RabbitLBJ]

namespace Solution3
{
	template<class T>
	struct CompareDeref
	{
		bool operator()(const T& a,const T& b) const 
		{
			return *a < *b;
		}
	};

	template<class T,class U>
	struct Pair2nd
	{
		const U& operator()(const pair<T,U>& a) const
		{
			return a.second;
		}
	};

	template<class IterIn,class IterOut>
	void sort_idxtbl(IterIn first,IterIn last,IterOut out)
	{
		multimap<IterIn,int,CompareDeref<IterIn> > v;
		for(int i = 0;first != last;++i,++first)
		{
			v.insert(make_pair(first,i));
		}
		transform(v.begin(),v.end(),out,Pair2nd<IterIn,int>());
	}
}


int main()
{
	int ai[10] = {15,12,13,14,18,11,10,17,16,19};
	cout << "----------------------" << endl;
	vector<int> vecai(ai,ai + 10);
	int aidxtbl[10];
	Solution3::sort_idxtbl(vecai.begin(),vecai.end(),aidxtbl);
	for(int i = 0;i < 10;i++)
	{
		cout << "i=" << i
			 << ",aidxtbl[i]=" << aidxtbl[i]
			 << ",ai[aidxtbl[i]]=" << ai[aidxtbl[i]]
			 << endl;
	}
	cout << "----------------------" << endl;
	cin.get();
	return 0;
}

[hen_hao_ji]

#include <iostream>
#include <algorithm>
using namespace std;

struct node
{
	int num;
	int index;
}r[100];
bool cmp(node x, node y)
{
	return x.num < y.num ? true:false;
}
int main()
{
	int a[]={1, 1, 4, 2, 3, 2, 1};

	int n = sizeof(a)/sizeof(int);
	
	for (int i=0; i < n; ++i)
	{
		r[i].num = a[i];
		r[i].index = i;
	}
	sort(r, r + n, cmp);
	
	return 0;
}