这是本人在研究leetcode中Median of Two Sorted Arrays一题目的时候看到一篇文章，觉得非常好，其中对快速排序重新实现。
文章来源于http://www.geeksforgeeks.org/这个网站。
We recommend to read following post as a prerequisite of this post.
K’th Smallest/Largest Element in Unsorted Array | Set 1
Given an array and a number k where k is smaller than size of array, we need to find the k’th smallest element in the given array. It is given that ll array elements are distinct.
Examples:
Input: arr[] = {7, 10, 4, 3, 20, 15}
k = 3
Output: 7
Input: arr[] = {7, 10, 4, 3, 20, 15}
k = 4
Output: 10
We have discussed three different solutions here.
In this post method 4 is discussed which is mainly an extension of method 3 (QuickSelect) discussed in the previous post. The idea is to randomly pick a pivot element. To implement randomized partition, we use a random function, rand() to generate index between l and r, swap the element at randomly generated index with the last element, and finally call the standard partition process which uses last element as pivot.
Following is C++ implementation of above Randomized QuickSelect.
// C++ implementation of randomized quickSelect#include<iostream>#include<climits>#include<cstdlib>using
namespace
std;int
randomPartition(int
arr[], int
l, int
r);// This function returns k‘th smallest element in arr[l..r] using// QuickSort based method. ASSUMPTION: ALL ELEMENTS IN ARR[] ARE DISTINCTint
kthSmallest(int
arr[], int
l, int
r, int
k){ // If k is smaller than number of elements in array if
(k > 0 && k <= r - l + 1) { // Partition the array around a random element and // get position of pivot element in sorted array int
pos = randomPartition(arr, l, r); // If position is same as k if
(pos-l == k-1) return
arr[pos]; if
(pos-l > k-1) // If position is more, recur for left subarray return
kthSmallest(arr, l, pos-1, k); // Else recur for right subarray return
kthSmallest(arr, pos+1, r, k-pos+l-1); } // If k is more than number of elements in array return
INT_MAX;}void
swap(int
*a, int
*b){ int
temp = *a; *a = *b; *b = temp;}// Standard partition process of QuickSort(). It considers the last// element as pivot and moves all smaller element to left of it and// greater elements to right. This function is used by randomPartition()int
partition(int
arr[], int
l, int
r){ int
x = arr[r], i = l; for
(int
j = l; j <= r - 1; j++) { if
(arr[j] <= x) { swap(&arr[i], &arr[j]); i++; } } swap(&arr[i], &arr[r]); return
i;}// Picks a random pivot element between l and r and partitions// arr[l..r] arount the randomly picked element using partition()int
randomPartition(int
arr[], int
l, int
r){ int
n = r-l+1; int
pivot = rand() % n; swap(&arr[l + pivot], &arr[r]); return
partition(arr, l, r);}// Driver program to test above methodsint
main(){ int
arr[] = {12, 3, 5, 7, 4, 19, 26}; int
n = sizeof(arr)/sizeof(arr[0]), k = 3; cout <<
"K‘th smallest element is "
<< kthSmallest(arr, 0, n-1, k); return
0;} |
Output:
K‘th smallest element is 5
Time Complexity:
The worst case time complexity of the above solution is still O(n2). In worst case, the randomized function may always pick a corner element. The expected time complexity of above randomized QuickSelect is Θ(n), see CLRS
book or MIT
video lecture for proof. The assumption in the analysis is, random number generator is equally likely to generate any number in the input range.
Sources:
MIT
Video Lecture on Order Statistics, Median
Introduction to Algorithms by Clifford Stein, Thomas
H. Cormen, Charles E. Leiserson, Ronald L.
This article is contributed by Shivam. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
good article————K’th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time)
原文地址：http://blog.csdn.net/chenxun_2010/article/details/42551269
