Brief： the-ith-element，given a array A with n element , return the i-th element of A.  A(n,i)

this problem can be solved using quick-sort idear, every time we choose a pivot ,after rearrangement, we know the pivot is the i-th element of A, so we can fixed this problem as the follow action.

i) choose  pivot, rearrangement array, get the pivot index of j;

ii) if j > i , then recursion the 1-th partition , A(1-th , i);

iii)if j < i, then recursion the 2-th partition, A(2-th, i-j);

 

 

int element(int *array, int left, int right, int index){	if(left >= right)		return array[left];	int pivot = array[left];	int i=0, j = 0;	for(i=left+1,j = left+1; j <=right; ++j)	{		if(array[j] < pivot)			swap(array[i++], array[j]);	}	swap(array[left], array[i-1]);	//pivot is the (i-left)-th element, and pivot's index is i-1	if(index == i-left)		return array[i-1];	else if(index < i-left)		return element(array, left, i-2, index);	else		return element(array, i,right, index-i+left);}