问题
I am currently studying quicksort and would like to know how it works when the first (or last) element is chosen as the pivot point.
Say for example I have the following array:
{15, 19, 34, 41, 27, 13, 9, 11, 44}
This is what I think happens:
{15, 19, 34, 41, 27, 13, 9, 11, 44}
^
pivot
{15, 19, 34, 41, 27, 13, 9, 11, 44}
^ ^
compare these two, they are good
{15, 19, 34, 41, 27, 13, 9, 11, 44}
^ ^
compare these two and swap
{11, 19, 34, 41, 27, 13, 9, 15, 44}
^ ^
compare these two and swap
{9, 19, 34, 41, 27, 13, 11, 15, 44}
^ ^
compare these two, they are good
{9, 19, 34, 41, 27, 13, 11, 15, 44}
^ ^
compare these two, they are good
{9, 19, 34, 41, 27, 13, 11, 15, 44}
^ ^
compare these two, they are good
{9, 19, 34, 41, 27, 13, 11, 15, 44}
^ ^
compare these two, they are good
{9, 19, 34, 41, 27, 13, 11, 15, 44}
^ ^
compare these two, they are good
{9, 19, 34, 41, 27, 13, 11, 15, 44}
End of first partition
Is this how it works? If so, would 19 be the new pivot point, or do you divide the array in half to find it (so that it would be 27/13), or does it depend on the implementation of the quicksort? Thanks for your time!
回答1:
Check wikipedia, there is a little example with a bit smaller list of inplace quicksort http://en.wikipedia.org/wiki/Quicksort
With your example the idea is to partition
{15, 19, 34, 41, 27, 13, 9, 11, 44}
into
{13, 9, 11 -- 15 -- 19, 34, 41, 27, 44}
So first we move pivot to the end
Swap 44, and 15
{44, 19, 34, 41, 27, 13, 9, 11, 15}
^ ^
Than check 44, its larger than pivot, so swap with one one before last...
{11, 19, 34, 41, 27, 13, 9, 44, 15}
^ ^
than check element at some position as last one was larger than pivot.
9 < 15, so proceed to the next, 19 > 15 => swap
{11, 9, 34, 41, 27, 13, 19, 44, 15}
^ ^
swap again
{11, 9, 13, 41, 27, 34, 19, 44, 15}
^ ^
next
{11, 9, 13, 41, 27, 34, 19, 44, 15}
^ ^
and second last swap
{11, 9, 13, 27, 41, 34, 19, 44, 15}
^
Now as forward and backward indices reached each other,
we swap pivot into right position
{11, 9, 13, 15, 41, 34, 19, 44, 27}
And we got partitioned set. Items less than 15 at the beginning, than pivot = 15, and then greater elements.
EDIT: algorithm described in wikipedia article is a bit different:
Legend:
^ = storeindex
# = i
{44, 19, 34, 41, 27, 13, 9, 11, 15}
^#
{44, 19, 34, 41, 27, 13, 9, 11, 15}
^ #
... until ...
{44, 19, 34, 41, 27, 13, 9, 11, 15}
^ #
{13, 19, 34, 41, 27, 44, 9, 11, 15}
^ #
{13, 9, 34, 41, 27, 44, 19, 11, 15}
^ #
{13, 9, 11, 41, 27, 44, 19, 34, 15}
^ #
{13, 9, 11, 15, 27, 44, 19, 34, 41}
^- pivot
回答2:
https://www.youtube.com/watch?v=COk73cpQbFQ for last element as pivot
int partition(int *a,int start,int end)
{
int pivot=a[end],temp,p1=start,i;
for(i=start;i<end;i++)
{
if(a[i]<pivot)
{
if(i!=p1)
{
temp=a[p1];
a[p1]=a[i];
a[i]=temp;
}
p1++;
}
}
temp=a[p1];
a[p1]=a[end];
a[end]=temp;
return p1;
}
https://www.youtube.com/watch?v=6UHCG64tHgo for first element as pivot
int partition1(int *a,int start,int end)
{
int pivot=a[start],p1=start+1,i,temp;
for(i=start+1;i<=end;i++)
{
if(a[i]<pivot)
{
if(i!=p1)
{
temp=a[p1];
a[p1]=a[i];
a[i]=temp;
} p1++;
}
}
a[start]=a[p1-1];
a[p1-1]=pivot;
return p1-1;
}
void quicksort(int *a,int start,int end)
{
int p1;
if(start<end)
{
p1=partition(a,start,end);
quicksort(a,start,p1-1);
quicksort(a,p1+1,end);
}
}
回答3:
/* Quick Sort taking first element as pivot element*/
void QuickSort(int* arr,int start,int last)
{
int i=start+1,j=last,temp;
if(i>j)
return;
while(i<=j)
{
if(arr[i]<arr[start])
{enter code here
i++;
}
if(arr[j]>arr[start])
{
j--;
}
if(i<=j)
{
temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;
}
}
temp=arr[start];
arr[start]=arr[j];
arr[j]=temp;
QuickSort(arr,start,j-1);
QuickSort(arr,j+1,last);
}
for whole code visit:-http://www.liladharpaliwal.blogspot.in/
回答4:
Choosing the first element as a pivot...
class QuickSortPart1{
public int partition(int[] a, int left, int right) {
int pivot = a[left];
while(left<=right) {
while(a[left] < pivot)
left++;
while(a[right] > pivot)
right--;
if(left<=right) {
int tmp = a[left];
a[left] = a[right];
a[right] = tmp;
left++;
right--;
}
}
return left;
}
public void recursiveQuickSort(int[] a, int i, int j) {
int idx = partition(a, i, j);
if(i < idx-1) {
recursiveQuickSort(a, i, idx-1);
}
if(j > idx) {
recursiveQuickSort(a, idx, j);
}
}
void printarray(int arr[]){
int len = arr.length;
for(int i=0; i<len; i++)
System.out.print(arr[i]+" ");
}
public static void main(String[] args) {
int arr[] = new int[]{5,8,1,3,7,9,2};
System.out.print(arr[i]+" ");
System.out.println("\n");
QuickSortPart1 ob = new QuickSortPart1();
ob.recursiveQuickSort(arr, 0, arr.length-1);
ob.printarray(arr);
}
}
回答5:
Try this algo: https://www.youtube.com/watch?v=7h1s2SojIRw
Base idea: All elements to left are smaller than all elements to the right then the element is said to be in 'sorted' position.
Algo:
// partition
Partition(l,h) {
pivot = A[l];
i=l; j=h;
while(i<j) {
do {
i++;
} while(A[i]<=pivot);
do {
j--;
} while(A[j]>pivot);
if(i<j) {
swap(i,j);
}
}
swap(A[l], A[j]);
return j;
}
// quicksort
QuickSort(l,h) {
pi = Partition(l, h);
QuickSort(l, pi);
QuickSort(pi+1, h);
}
回答6:
the following code uses first element as pivot
public static int[] qs(int[] list, int start, int end) {
if (end - start <= 1) {
return list;
}
int pivot = split(list, start, end);
qs(list, start, pivot);
qs(list, pivot + 1, end);
return list;
}
private static int split(int[] list, int start, int end) {
int pivot = list[start];
int i = start;
for (int j = start + 1; j <= end; j++) {
int current = list[j];
if (current < pivot) {
swap(list, i + 1, j);
i++;
}
}
swap(list, start, i);
return i;
}
private static void swap(int[] list, int i, int j) {
int temp = list[i];
list[i] = list[j];
list[j] = temp;
}
来源：https://stackoverflow.com/questions/6740183/quicksort-with-first-element-as-pivot-example