LeetCode OJ 215. Kth Largest Element in an Array 堆排序求解
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题目链接:https://leetcode.com/problems/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.
Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.
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求给定数组的第K大的元素。如果先排序,然后取第K个元素,那么时间复杂度是O(n*log n)。借助堆的数据结构,可以把时间复杂度降到O(n*logk)。
如果求第K大的元素,那么要构建的是小顶堆。求第K小的元素,那么要构建大顶堆!先构建k个元素的堆,另外i-k个元素逐个跟堆顶元素比较,如果比堆顶元素小,那么就将该元素纳入堆中,并保持堆的性质。
我的AC代码
public class KthLargestElementinanArray { public static void main(String[] args) { int[] a = { 3, 2, 1, 5, 6, 4 }; System.out.println(findKthLargest(a, 1)); int[] b = { -1,2,0}; System.out.println(findKthLargest(b, 3)); int[] c = { 3,1,2,4}; System.out.println(findKthLargest(c, 2)); } public static int findKthLargest(int[] nums, int k) { int[] heap = new int[k]; heap[0] = nums[0]; for (int i = 1; i < k; i++) { siftUp(nums[i], heap, i); } for (int i = k; i < nums.length; i++) { siftDown(nums, k, heap, i); } return heap[0]; } private static void siftDown(int[] nums, int k, int[] heap, int i) { if (nums[i] > heap[0]) { heap[0] = nums[i]; int p = 0; while(p < k) { int minChild = 2 * p + 1; if(minChild + 1 < k && heap[minChild] > heap[minChild + 1]) minChild ++; if(minChild < k && heap[p] > heap[minChild]) { swap(heap, p, minChild); p = minChild; } else break; } } } private static void siftUp(int num, int[] heap, int i) { int p = i; heap[i] = num; while (p != 0) { int parent = (p - 1) / 2; if (heap[parent] > heap[p]) { swap(heap, p, parent); } p = parent; } } private static void swap(int[] heap, int p, int parent) { int temp = heap[parent]; heap[parent] = heap[p]; heap[p] = temp; } }
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