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
1 public class Solution { 2 public int findKthLargest(int[] nums, int k) { 3 if (nums == null || k > nums.length) 4 return 0; 5 int length = nums.length, index = length - k, result = 0; 6 PriorityQueue<Integer> pq = new PriorityQueue<Integer>(length, 7 new Comparator<Integer>(){ 8 public int compare(Integer a, Integer b) { 9 return (a - b); 10 } 11 }); 12 for (int i = 0; i < length; i++) 13 pq.add(nums[i]); 14 while (index >= 0) { 15 result = pq.poll(); 16 index--; 17 } 18 return result; 19 } 20 }
Improvement
1 public class Solution { 2 public int findKthLargest(int[] nums, int k) { 3 if (nums == null || k > nums.length) 4 return 0; 5 int length = nums.length, index = k, result = 0; 6 PriorityQueue<Integer> pq = new PriorityQueue<Integer>(length, Collections.reverseOrder()); 7 for (int i = 0; i < length; i++) 8 pq.add(nums[i]); 9 while (index > 0) { 10 result = pq.poll(); 11 index--; 12 } 13 return result; 14 } 15 }
Solution 2 -- Quick Select
Average Time O(n)
1 ''' 2 a: [3, 2, 5, 1], 2 3 output: 2 4 ''' 5 6 def swap(a, index1, index2): 7 if index1 >= index2: 8 return 9 tmp = a[index1] 10 a[index1] = a[index2] 11 a[index2] = tmp 12 13 def partition(a, start, end): 14 ''' a[start:end] 15 pivot: a[end] 16 return: index of pivot 17 ''' 18 pivot = a[end] 19 cur_big = start - 1 20 for i in range(start, end): 21 if a[i] >= pivot: 22 cur_big += 1 23 swap(a, cur_big, i) 24 cur_big += 1 25 swap(a, cur_big, end) 26 return cur_big 27 28 29 def quick_select(a, k): 30 k -= 1 31 length = len(a) 32 start = 0 33 end = length - 1 34 while start < end: 35 index = partition(a, start, end) 36 if index == k: 37 return a[index] 38 if index > k: 39 # Note: end != index 40 end = index - 1 41 else: 42 # Note: start != index 43 start = index + 1 44 k = k - index 45 return -1 46 47 a = [3, 2, 5, 1] 48 k = 1 49 print(quick_select(a, k))
