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将k值当做物理地址的值，比如第5个数其实就是数组4的位置，第2个数就是数组1的位置
每次只需要判断k在左区间还是右区间，一直递归查找k所在区间
最后只剩一个数时，只会有数组[k]一个数，返回数组[k]的值就是答案描述将k值当做物理地址的值，比如第5个数其实就是数组4的位置，第2个数就是数组1的位置每次只需要判断k在左区间还是右区间，一直递归查找k所在区间最后只剩一个数时，只会有数组[k]一个数，返回数组[k]的值就是答案import java.util.Scanner;public class Main{ static int N = 100010; static int[] A = new int[N]; static int n, k; public static void main(String[] args){ Scanner sc = new Scanner(System.in); n = sc.nextInt(); k = sc.nextInt(); for(int i = 0; i < n; i++) A[i] = sc.nextInt(); System.out.println(quickSort(0, n-1, k-1)); } public static int quickSort(int l, int r, int k){ if(l >= r) return A[k]; int x = A[l], i = l-1, j = r+1; while(i < j) { do i++; while(A[i] < x); do j--; while(A[j] > x); if(i < j) { int temp = A[i]; A[i] = A[j]; A[j] = temp; } } if(k <= j) return quickSort(l, j, k); else return quickSort(j+1, r, k); }} import java.util.*;class Main{ public static void main(String[] args){ Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int k = sc.nextInt(); int[] arr = new int[n]; for(int i = 0; i < n; i++){ arr[i] = sc.nextInt(); } quickSort(arr,0,n-1); System.out.print(arr[k-1]); } static void quickSort(int[] arr,int l,int r){ if(l>=r){ return; } int i = l-1; int j = r+1; int mid = arr[(l+r)/2]; while(i mid); if(i import java.util.*;//把整个数组建成堆，每次输出堆顶即可//因此只需要用到输出集合的最小值//还有删除最小值，关键是实现down操作。//求最小值是o1的时间复杂度，而求插入和删除的时间复杂度与树的高度有关因此是logn的class Main{ static int m,n,size; static int[] p = new int[100010]; public static void main(String [] args){ Scanner sc = new Scanner(System.in); n = sc.nextInt(); m = sc.nextInt(); for(int i = 1; i <=n; i++){ p[i] = sc.nextInt(); } size = n; for(int i = n/2; i !=0;i--){ down(i); } for(int i = 1; i <=m; i++){ if(i == m){ System.out.println(p[1]); } p[1] = p[size]; size--; down(1); } } static void down(int u){ int t = u; //如果子节点 import java.util.*;class Main{ public static void main(String []args){ Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int k = sc.nextInt(); int[] arr = new int[n]; for(int i = 0; i < n; i++){ arr[i] = sc.nextInt(); } Arrays.sort(arr); System.out.println(arr[k-1]); }} import java.util.PriorityQueue;import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int k = sc.nextInt(); PriorityQueue queue = new PriorityQueue<>(); for (int i = 0; i < n; i++){ int x = sc.nextInt(); queue.offer(x); } for(int i = 0; i < k-1; i++){ queue.poll(); } System.out.println(queue.peek()); }} 地址：
class Solution { public int findKthLargest(int[] nums, int k) { return quick_sort(nums, 0, nums.length - 1, k - 1); } int quick_sort(int[] nums, int l, int r, int k) { if (l == r) return nums[k]; int x = nums[l], i = l - 1, j = r + 1; while (i < j) { do i ++ ; while (nums[i] > x); do j -- ; while (nums[j] < x); if (i < j) { int tmp = nums[i]; nums[i] = nums[j]; nums[j] = tmp; } } if (k <= j) return quick_sort(nums, l, j, k); else return quick_sort(nums, j + 1, r, k); } } class Solution { public int findKthLargest(int[] nums, int k) { quick_sort(nums, 0, nums.length - 1); return nums[k-1]; } void quick_sort(int[] nums, int l, int r) { if (l >= r) return; int x = nums[l], i = l - 1, j = r + 1; while (i < j) { do i ++ ; while (nums[i] > x); do j -- ; while (nums[j] < x); if (i < j) { int tmp = nums[i]; nums[i] = nums[j]; nums[j] = tmp; } } quick_sort(nums, l, j); quick_sort(nums, j + 1, r); }} class Solution { public int findKthLargest(int[] nums, int k) { int[] arr = new int[nums.length+1]; for(int i = 1; i <= nums.length; i++){ arr[i] = nums[i-1]; } int n = nums.length; int size = n; for(int i = n/2; i !=0;i--){ down(i,arr,size); } for(int i = 1; i <= k; i++){ if(i == k){ return arr[1]; } arr[1] = arr[size]; size--; down(1,arr,size); } return 0; } void down(int u,int[] arr,int size){ int t = u; //如果子节点 <当前节点的话就把下标位置赋给t if(u*2<="size&&arr[u*2]"> arr[t]){ t = u*2; } if(u*2+1<=size&&arr[u*2+1] > arr[t]){ t = u*2+1; } if(u!=t){ int temp = arr[u]; arr[u] = arr[t]; arr[t] = temp; down(t,arr,size); } } } class Solution { public int findKthLargest(int[] nums, int k) { Arrays.sort(nums); for(int i = nums.length -1; i >= 0; i--){ if(i == nums.length-k){ return nums[i]; } } return 0; } } class Solution { public int findKthLargest(int[] nums, int k) { PriorityQueue queue = new PriorityQueue (((o1, o2) -> { return o2-o1; })); for (int i = 0; i < nums.length; i++){ queue.offer(nums[i]); } for(int i = 0; i < k-1; i++){ queue.poll(); } return queue.peek(); } } 转载地址：http://harwi.baihongyu.com/