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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;
}
}
LeetCode OJ 215. Kth Largest Element in an Array 堆排序求解
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原文地址：http://blog.csdn.net/bruce128/article/details/50722572
