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.
本题开始我用了数组排序的方法做，代码如下：
public class Solution {
public int findKthLargest(int[] nums, int k) {
Arrays.sort(nums);
int kth = 0;
for(int i=nums.length-1;i>=0;i--){
k--;
if(k==0) kth = nums[i];
}
return kth;
}
}
后来我又用了大顶堆的概念来做，代码如下：
public class Solution {
public int findKthLargest(int[] nums, int k) {
PriorityQueue<Integer> q = new PriorityQueue<Integer>(nums.length,new Comparator<Integer>(){
public int compare(Integer a,Integer b){
return b-a;
}
});
for(int i:nums){
q.offer(i);
}
k--;
while(k>0){
q.poll();
k--;
}
return q.peek();
}
}
大神用了快速排序法来做，很聪明的解题办法，代码如下：
public class Solution {
public int findKthLargest(int[] nums, int k) {
int n = nums.length;
int q = quickselect(nums,0,n-1,n-k+1);
return nums[q];
}
public void swap(int[] nums,int i,int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
public int quickselect(int[] nums,int lo,int hi,int k){
int pivot = nums[hi];
int i = lo;
int j = hi;
while(i<j){
if(nums[i++]>pivot) swap(nums,--i,--j);
}
swap(nums,i,hi);
int order = i-lo+1;
if(order==k) return i;
else if(order>k) return quickselect(nums,lo,i-1,k);
else return quickselect(nums,i+1,hi,k-order);
}
}