Kth Largest Element in an Array 解答

Question

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.

Solution 1 -- PriorityQueue

1. 将所有元素放入heap中

2. 依次用poll()取出heap中最大值

 public class Solution {
     public int findKthLargest(int[] nums, int k) {
         if (nums == null || k > nums.length)
             return 0;
         int length = nums.length, index = length - k, result = 0;
         PriorityQueue<Integer> pq = new PriorityQueue<Integer>(length,
                                                             new Comparator<Integer>(){
                                                                 public int compare(Integer a, Integer b) {
                                                                     return (a - b);
                                                                 }
                                                             });
         for (int i = 0; i < length; i++)
             pq.add(nums[i]);
         while (index >= 0) {
             result = pq.poll();
             index--;
         }
         return result;
     }
 }

Improvement

 public class Solution {
     public int findKthLargest(int[] nums, int k) {
         if (nums == null || k > nums.length)
             return 0;
         int length = nums.length, index = k, result = 0;
         PriorityQueue<Integer> pq = new PriorityQueue<Integer>(length, Collections.reverseOrder());
         for (int i = 0; i < length; i++)
             pq.add(nums[i]);
         while (index > 0) {
             result = pq.poll();
             index--;
         }
         return result;
     }
 }

Solution 2 -- Quick Select

Quick Sort in Java

Average Time O(n)

 '''
 a: [3, 2, 5, 1], 2
 output: 2
 '''

 def swap(a, index1, index2):
     if index1 >= index2:
         return
     tmp = a[index1]
     a[index1] = a[index2]
     a[index2] = tmp

 def partition(a, start, end):
     ''' a[start:end]
         pivot: a[end]
         return: index of pivot
     '''
     pivot = a[end]
     cur_big = start - 1
     for i in range(start, end):
         if a[i] >= pivot:
             cur_big += 1
             swap(a, cur_big, i)
     cur_big += 1
     swap(a, cur_big, end)
     return cur_big

 def quick_select(a, k):
     k -= 1
     length = len(a)
     start = 0
     end = length - 1
     while start < end:
         index = partition(a, start, end)
         if index == k:
             return a[index]
         if index > k:
             # Note: end != index
             end = index - 1
         else:
             # Note: start != index
             start = index + 1
             k = k - index
     return -1

 a = [3, 2, 5, 1]
 k = 1
 print(quick_select(a, k))