Determine the positioning of one random pivot each time
Place each element to both sides of pivot
Repeat for both sides of pivot
Time Comp:
On average: O(N logN)
Ideally, each call splits the array into left and right of pivot. Calling logN times.
Worst Case: O(N^2)
If you have terrible karma, the pivot might be chosen on one of the ends every time. Ex: 012345 pivot = 0 , 012345 pivot = 1, 012345 pivot = 2 .....
Every element is pivot, one at a time, thus n^2
Space Comp: O(logN) - O(N) call stacks
public class Solution {
private void swap (int[] array, int i, int j){
int tmp = array[i];
array[i] = array[j];
array[j] = tmp;
}
public int[] quickSort(int[] array) {
if (array == null) return null;
quickSortHelper(array, 0, array.length - 1);
return array;
}
public void quickSortHelper(int[] array, int left, int right){
//base case: one element array
if (left >= right){
return;
}
//choose the pivot
int pivot = partition(array, left, right); //sorts one element
quickSortHelper(array, left, pivot - 1);
quickSortHelper(array, pivot + 1, right);
}
public int partition(int[] array, int left, int right) {
int pivotIndex = pivotIndex(left, right);
int pivot = array[pivotIndex];
swap ( array, pivotIndex, right);
int leftBound = left;
int rightBound = right - 1; //excluding pivot
while( leftBound <= rightBound) {
if (array[leftBound] < pivot) {
leftBound++; //if l < p, correct placing, incur next
} else if (array[rightBound] >= pivot) {
rightBound--; //if r >= p, correct placing, incur next
} else {
swap(array, leftBound++, rightBound--); //else both in wrong place, swap
}
}
swap(array, leftBound, right);
return leftBound; //leftBound ends at pivot
}
public int pivotIndex(int left, int right) {
return left + (int) (Math.random() * (right - left + 1));
}
}
Mistake to watch out for:
after partition, new position of pivot value is on leftBound , not on pivotIndex.