Leetcode: Kth Largest Element in an Array

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.For example,
Given [3,2,1,5,6,4] and k = 2, return 5.Note: 
You may assume k is always valid, 1 ≤ k ≤ array's length.

The same with Lintcode: Kth largest element

快速选择 Quick Select

复杂度

时间 Avg O(N) Worst O(N^2) 空间 O(1)

One thing to notice is 23-24 line, still find kth smallest element, do not need to decrease from k because the indexing(编号) does not change

 1 public class Solution {
 2     public int findKthLargest(int[] nums, int k) {
 3         int len = nums.length;
 4         return findKthSmallest(nums, 0, len-1, len-k+1);
 5     }
 6     
 7     public int findKthSmallest(int[] nums, int start, int end, int k) {
 8         int l = start;
 9         int r = end;
10         int pivot = end;
11         while (l < r) {
12             while (l<r && nums[l] < nums[pivot]) {
13                 l++;
14             }
15             while (l<r && nums[r] >= nums[pivot]) {
16                 r--;
17             }
18             if (l == r) break;
19             swap(nums, l, r);
20         }
21         swap(nums, l, pivot);
22         if (l+1 == k) return nums[l];
23         else if (l+1 < k) return findKthSmallest(nums, l+1, end, k);
24         else return findKthSmallest(nums, start, l-1, k);
25     }
26     
27     public void swap(int[] nums, int l, int r) {
28         int temp = nums[l];
29         nums[l] = nums[r];
30         nums[r] = temp;
31     }
32 }

优先队列

复杂度

时间 O(NlogK) 空间 O(K)

思路

遍历数组时将数字加入优先队列（堆），一旦堆的大小大于k就将堆顶元素去除，确保堆的大小为k。遍历完后堆顶就是返回值。

 1 public class Solution {
 2     public int findKthLargest(int[] nums, int k) {
 3         PriorityQueue<Integer> p = new PriorityQueue<Integer>();
 4         for(int i = 0 ; i < nums.length; i++){
 5             p.add(nums[i]);
 6             if(p.size()>k) p.poll();
 7         }
 8         return p.poll();
 9     }
10 }