方法一:排序,计数器
class Solution {
public int findKthLargest(int[] nums, int k) {
Arrays.sort(nums);
return nums[nums.length-k];
}
}
方法二: 用堆
最小堆 + 维护一个只有k个元素;
class Solution {
public int findKthLargest(int[] nums, int k) {
//创建堆,注意这里应用了lambda表达式
PriorityQueue<Integer> heap = new PriorityQueue<>((n1,n2) ->(n1-n2));
for(int i :nums){
heap.add(i);
//维持当前堆的大小保持在k个元素
if(heap.size()>k)
heap.poll();
}
return heap.poll();
}
}
方法三: 快速选择排序算法(注意不是快速排序,是基于快排修改得来的)
import java.util.Random;
class Solution {
int [] nums;
public void swap(int a, int b) {
int tmp = this.nums[a];
this.nums[a] = this.nums[b];
this.nums[b] = tmp;
}
//划分
public int partition(int left, int right, int pivot_index) {
int pivot = this.nums[pivot_index];
// 1. move pivot to end
swap(pivot_index, right);
int store_index = left;
// 2. move all smaller elements to the left
for (int i = left; i <= right; i++) {
if (this.nums[i] < pivot) {
swap(store_index, i);
store_index++;
}
}
// 3. move pivot to its final place
swap(store_index, right);
return store_index;
}
public int quickselect(int left, int right, int k_smallest) {
/*
Returns the k-th smallest element of list within left..right.
*/
if (left == right) // If the list contains only one element,
return this.nums[left]; // return that element
// select a random pivot_index
Random random_num = new Random();
int pivot_index = left + random_num.nextInt(right - left);
pivot_index = partition(left, right, pivot_index);
// the pivot is on (N - k)th smallest position
if (k_smallest == pivot_index)
return this.nums[k_smallest];
// go left side
else if (k_smallest < pivot_index)
return quickselect(left, pivot_index - 1, k_smallest);
// go right side
return quickselect(pivot_index + 1, right, k_smallest);
}
public int findKthLargest(int[] nums, int k) {
this.nums = nums;
int size = nums.length;
// kth largest is (N - k)th smallest
return quickselect(0, size - 1, size - k);
}
}