smallValue = 7; int[] quicksort(int[] A) { if (A.length < smallValue) return insertionSort(A); // Base case of the recursion p = a pivot for A, a number in A that is more or less in the middle of the values of A; B1 = an array with the numbers less than p; B2 = an array with the numbers greater than p; C1 = quicksort(B1); // Recursively solve on the subproblems C2 = quicksort(B2); D = C1 followed by p followed by C2 // Combine return D; }We'll come back later to the question of how the pivot is chosen. In the example below, I'll just pick it "magically" to be some number more or less in the middle.

As with mergesort, when you get down to arrays of size 7 or so, you just turn to an insertion sort for efficiency, rather than recurring down to size 1.

A = [31,41,59,26,53,58,27,18,28,45,9,14,42,12,17] quicksort(A) pivot = 31; B1 = [26,27,18,28,9,14,12,17] B2 = [41,59,53,58,45,42] C1 = quicksort([26,27,18,28,9,14,12,17] pivot = 17; B1 = [9,14,12] B2 = [26,27,28] C1=quicksort([9,14,12]) = [9,12,14] C2=quicksort([26,27,18,28]) = [18,26,27,28] D = append(C1,[pivot],C2) = [9,12,14,17,18,26,27,28] C1 = [9,12,14,17,18,26,27,28] C2 = quicksort([41,59,53,58,45,42]) pivot = 53; B1 = [41,45,42] B2 = [59,58] C1 = quicksort([41,45,42]) = [41,42,45] C2 = quicksort([59,58]) = [58,59] D = append(C1,[pivot],C2) = [41,42,45,53,58,59]

To do this, it uses a two-fingered method. First, move the pivot itself to the front of the array, to get it out of the way. Then run the left finger up from the left, until you find a number larger than the pivot. Run the right finger from the right, until you find a number smaller than the pivot. Swap these. Continue on, until the two fingers meet. Then put the pivot into its right place. This is called the "partition" function. (There are many different ways of writing the partition function, none of them quite as elegant as one would like.)

// Partition the elements of array A between indices L and U inclusive, // using the value initially at A[IPIVOT] partition(a,l,u,ipivot) { pivot = a[ipivot]; swap(a[ipivot],a[l]); // Move the pivot to the front. i = l+1; // left-handed pointer. j = u; // right-handed pointer. while (i < j) { while (i < j && a[i] < pivot) i++; while (j > i && a[j] > pivot) j-- if (i < j) { swap(a[i],a[j]); i++; j--; } } // the partitioning is complete. if (a[i] < pivot) ipivot = i; // i is the last small number else ipivot = i-1; // i-1 is the last small number. swap(a[l],a[ipivot]) // swap the pivot into the middle position. return ipivot; } // return the new position of the pivot.

A = [31,41,59,26,53,58,27,18,28,45,9,14,42,12,17] Choose ipivot = 1 (pivot = 41). partition(A,0,14,2) swap(A[ipivot],A[0]); A = [41,31,59,26,53,58,27,18,28,45,9,14,42,12,17] i = 1; j = 14; move i up to 2 (A[i]=59 > pivot) keep j at 14 (A[i] = 17 < pivot) swap(A[i],A[j]) A = [41,31,17,26,53,58,27,18,28,45,9,14,42,12,59] i j Move i up to 4 (A[i] = 53 > pivot) Move j back to 12 (A[i] = 12 < pivot) swap(A[i],A[j]) A = [41,31,17,26,12,58,27,18,28,45,9,14,42,53,59] i j Move i up to 5 (A[i]=58 > pivot) Move j back to 11 (A[i] = 14 < pivot) swap(A[i],A[j]) A = [41,31,17,26,12,14,27,18,28,45,9,58,42,53,59] i j Move i up to 9 (A[i]=45 > pivot) Move j back to 10 (A[i] = 14 < pivot) swap(A[i],A[j]) A = [41,31,17,26,12,14,27,18,28,9,45,58,42,53,59] i j i++ = 10, j-- = 9, so the outer loop exits ipivot = 9; swap(A[l],A[ipivot] A = [9,31,17,26,12,14,27,18,28,41,45,58,42,53,59] return ipivot = 9, the index of the pivot. Note that everything before 9 is less than the pivot and everything after is greater.

smallValue = 7; // quicksort array A between indices l and u int[] quicksort(int[] A, int l,u) { if (l - u < smallValue) return insertionSort(A); // Base case of the recursion ipivot = choose an index for a pivot; m = partition(A,l,u,ipivot); // Divide into small and large numbers quicksort(A,l,m-1); // Recursively sort the small numbers quicksort(A,m+1,u); // Recursively sort the large numbers } // the combination step is nothing, because the numbers are // already where they should be.

A = [31,41,59,26,53,58,27,18,28,45,9,14,42,12,17] qs(0,14) pivot = 31 partition(0,14,0) A = [[28,17,12,26,14,9,27,18] 31 [43,58,53,42,59,41]] partition returns 8 qs(0,7) pivot = 28 partition(0,7,0) A = [[[18,17,12,26,14,9,27], 28] 31 [43,58,53,42,59,41]] partition returns 6 qs(0,6) pivot = 18 partition(0,6,0) A = [[[[14,7,12,9], 18, [26,27]],28],31,[43,58,53,42,59,41]] partition returns 4 qs(0,3) pivot=14 partition(0,3,0) A = [[[[[9,7,12],14], 18, [26,27],28],31,[43,58,53,42,59,41]] partition returns 2 qs(0,2) calls insertion sort A = [[[[[7,9,12],14], 18, [26,27],28],31,[43,58,53,42,59,41]] qs(0,2) returns qs(0,3) returns qs(5,6) calls insertion sort A = [[7,9,12,14,18,26,27,28],31,[43,58,53,42,59,41]] qs(5,6) returns qs(0,6) returns qs(0,7) returns qs(9,14) pivot = 43 partition(9,14,9) A = [[7,9,12,14,18,26,27,28],31,[[42,41],43,[53,59,58]]] partition returns 11 qs(9,10) calls insertion sort A = [[7,9,12,14,18,26,27,28],31,[[41,42],43,[53,59,58]]] qs(9,10) returns qs(12,14) calls insertion sort A = [[7,9,12,14,18,26,27,28],31,[[41,42],43,[53,58,59]]] qs(12,14) returns qs(9,14) returns qs(0,14) returns A = [[7,9,12,14,18,26,27,28],31,[[41,42],43,[53,58,59]]]

The running time of quicksort depends on whether the pivot actually divides the data more or less in half.

You can show by a similar argument that if there is a fixed maximum "lopsidedness" of the partition --- for instance, if the division is never worse than 90% on one side and 10% on the other ---- then the overall running time is still O(N*log(N)), though the constant factor is larger.

- You pick the pivot at random.
- You pick the pivot systematically (e.g. A[0]) but the initial order of the data is random.

For practical purposes, you can't.

In theory, there is an O(N) algorithm to find the median element of an array, and if you use that to choose the pivot, you get an O(N * log(N)) worst case version of quicksort. However, the median algorithm is so complicated that this version of quicksort will run way slower than heapsort or mergesort. So there's no point to it.

A couple of strategies people use in practice.

- Take the first, the last, and the middle element and take the median of those as the pivot. This is likely to be reasonaby close to the median if the data is random, and likely to be the median if the data is originally sorted or nearly sorted.
- Take a random element.
- Take K random elements for some small K and use the median of those.