Find K-th largest element in an array. and N is much larger than k. Note that it is the kth largest element in the sorted order, not the kth distinct element.
Example
Example 1:
Input:[9,3,2,4,8],3
Output:4
Example 2:
Input:[1,2,3,4,5,6,8,9,10,7],10
Output:1
Notice
You can swap elements in the array
思路：可以用quick select模板,注意模板是两个while一个if，另外条件是i <= j
public class Solution {
/**
* @param nums: an integer unsorted array
* @param k: an integer from 1 to n
* @return: the kth largest element
*/
public int kthLargestElement2(int[] nums, int k) {
if(nums == null || nums.length == 0) {
return -1;
}
return findKth(nums, k, 0, nums.length-1);
}
private int findKth(int[] nums, int k, int start, int end) {
if(start == end) {
return nums[start];
}
int mid = start + (end - start) / 2;
int pivot = nums[mid];
int i = start; int j = end;
while(i <= j) {
while(i <= j && nums[i] > pivot) {
i++;
}
while(i <= j && nums[j] < pivot) {
j--;
}
if(i <= j) {
swap(nums, i, j);
i++;
j--;
}
}
if(start + k - 1 <= j) {
return findKth(nums, k, start, j);
}
if(start + k - 1 >= i) {
return findKth(nums, k - (i-start), i, end);
}
return nums[j+1];
}
private void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}思路2：如果按照题目意思，如果num很大很大的话，那么用priorityqueue，nlogk
public class Solution {
/**
* @param nums: an integer unsorted array
* @param k: an integer from 1 to n
* @return: the kth largest element
*/
public int kthLargestElement2(int[] nums, int k) {
if(nums == null || nums.length == 0) {
return -1;
}
PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
for(int i = 0; i < nums.length; i++) {
if(pq.size() < k) {
pq.add(nums[i]);
} else {
if(nums[i] > pq.peek()){
pq.poll();
pq.add(nums[i]);
}
}
}
return pq.peek();
}
}