Kth Largest Element
Find K-th largest element in an array.
Note
You can swap elements in the array
Example
In array [9,3,2,4,8], the 3th largest element is 4
Challenge
O(n) time, O(1) space
原题链接：
http://www.lintcode.com/en/problem/kth-largest-element/
SOLUTION 1:
使用改进的Quicksort partition，可以达到O(n)的时间复杂度，并且不需要额外空间。
请参考： http://en.wikipedia.org/wiki/Quickselect#Time_complexity
时间复杂度：如果我们是这样的数组：1，2，3，4，5，然后又每一次取左边的当pivot，就会达到最坏的时间复杂度。也就是O(N2)
我们也有一些解决方法：
以下来自维基，我们可以取3个数进行平均。
http://stackoverflow.com/questions/7559608/median-of-three-values-strategy
The easiest solution is to choose a random pivot, which yields almost certain linear time. Deterministically, one can use median-of-3 pivot strategy (as in the quicksort), which yields linear performance on partially sorted data, as is common in the real world. However, contrived sequences can still cause worst-case complexity; David Musser describes a "median-of-3 killer" sequence that allows an attack against that strategy, which was one motivation for his introselect algorithm.
1 package Algorithms.algorithm.NChapter.findKthNumber;
2
3 import java.util.ArrayList;
4
5 // The link:
6
7 // http://www.lintcode.com/en/problem/kth-largest-element/
8
9 class KthLargestElement_lintCode {
10 //param k : description of k
11 //param numbers : array of numbers
12 //return: description of return
13 public int kthLargestElement(int k, ArrayList<Integer> numbers) {
14 // write your code here
15 if (k < 1 || numbers == null) {
16 return 0;
17 }
18
19 return getKth(numbers.size() - k + 1, numbers, 0, numbers.size() - 1);
20 }
21
22 public int getKth(int k, ArrayList<Integer> numbers, int start, int end) {
23 // Choose the last one as the pivot
24 int pivot = numbers.get(end);
25
26 int left = start;
27 int right = end;
28
29 while (true) {
30 while (numbers.get(left) < pivot && left < right) {
31 left++;
32 }
33
34 while (numbers.get(right) >= pivot && right > left) {
35 right--;
36 }
37
38 if (left == right) {
39 break;
40 }
41
42 swap(numbers, left, right);
43 }
44
45 // left: the first one which is bigger than pivot.
46 swap(numbers, left, end);
47
48 if (k == left + 1) {
49 return pivot;
50 // Try to find the element from the left side.
51 } else if (k < left + 1) {
52 return getKth(k, numbers, start, left - 1);
53 } else {
54 // Try to find the element from the right side.
55 return getKth(k, numbers, left + 1, end);
56 }
57 }
58
59 /*
60 Swap the two nodes.
61 */
62 public void swap(ArrayList<Integer> numbers, int n1, int n2) {
63 int tmp = numbers.get(n1);
64 numbers.set(n1, numbers.get(n2));
65 numbers.set(n2, tmp);
66 }
67 };
SOLUTION 2:
以下这些链接有详细的讨论，就不再一一叙述了。目前来讲Solution 1应该是最优的啦。
http://www.geeksforgeeks.org/k-largestor-smallest-elements-in-an-array/
http://www.quora.com/What-is-the-most-efficient-algorithm-to-find-the-kth-smallest-element-in-an-array-having-n-elements
GITHUB: