Chapter 19: Problem 10
The recursive sorting technique called quicksort uses the following basic algorithm for a one-dimensional array of values: a) Partitioning Step: Take the first element of the unsorted array and determine its final lo- cation in the sorted array (i.e., all values to the left of the element in the array are less than the element, and all values to the right of the element in the array are greater than the element-we show how to do this below). We now have one element in its proper location and two unsorted subarrays. b) Rectorsive Step: Perform Step 1 on each unsorted subarray, Each time Step \(I\) is performed on a subarray, another element is placed in its final location of the sorted array, and two unsorted subarrays are created. When a subarray consists of one element, that element is in its final location (because a one-element array is already sorted). The basic algorithm seems simple enough, but how do we determine the final position of the first element of each subarray? As an example, consider the following set of values (the element in bold is the partitioning element- it will be placed in its final location in the sorted array): \(\begin{array}{llllllllll}37 & 2 & 6 & 4 & 89 & 8 & 10 & 12 & 68 & 45\end{array}\) Starting from the rightmost element of the array, compare each element with 37 until an element less than 37 is found; then swap 37 and that element. The first element less than 37 is \(12,\) so 37 and 12 are swapped. The new array is Starting from the right, but beginning with the element before \(89,\) compare each element with 37 until an element less than 37 is found - then swap 37 and that element. The first element less than 37 is \(10,\) so 37 and 10 are swapped. The new array is \\[ \begin{array}{llllllllll} 12 & 2 & 6 & 4 & 10 & 8 & 37 & 89 & 68 & 45 \end{array} \\] Starting from the left, but beginning with the element after \(10,\) compare each element with 37 until an element greater than 37 is found - then swap 37 and that element. There are no more elements greater than \(37,\) so when we compare 37 with itself, we know that 37 has been placed in its final location in the sorted array. Every value to the left of 37 is smaller than it, and every value to the right of 37 is larger than it. Once the partition has been applied on the previous array, there are two unsorted subarrays. The subarray with values less than 37 contains 12,2,6,4,10 and \(8 .\) The subarray with values greater than 37 contains 89,68 and \(45 .\) The sort continues recursively, with both subarrays being partitioned in the same manner as the original array. Based on the preceding discussion, write recursive method quicksortHelper to sort a one-dimensional integer array. The method should receive as arguments a starting index and an ending index on the original array being sorted.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.