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  • 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

    Quick Sort in Java

    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))
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  • 原文地址:https://www.cnblogs.com/ireneyanglan/p/4827845.html
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