Java 如何在O(1)空间中打印优先级队列的降序
我正在解决堆上的问题,我希望使用PriorityQueue按降序输出问题。Java 如何在O(1)空间中打印优先级队列的降序,java,arrays,heap,priority-queue,heapsort,Java,Arrays,Heap,Priority Queue,Heapsort,我正在解决堆上的问题,我希望使用PriorityQueue按降序输出问题。 输入: 一, 5.2 125787123 输出: 23787 想要的输出: 787 23 class GFG { public static void main(String args[])throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); int t = Integ
输入:
一,
5.2
125787123
输出:
23787
想要的输出:
787 23
class GFG {
public static void main(String args[])throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t-->0) {
// take array and kth element iput
String n_k[] = br.readLine().split(" ");
// store array size in n && kth element to find in k
int n = Integer.parseInt(n_k[0]);
int k = Integer.parseInt(n_k[1]);
// Array String input
String s[] = br.readLine().split(" ");
int d[] = new int[n];
for(int i = 0 ; i < n ; i++) {
d[i] = Integer.parseInt(s[i]);
}
PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>();
// delete the minimum element in the element and just keep
// k greater element in the minHeap
for(int i = 0 ; i < n; i++) {
minHeap.add(d[i]);
// if size of heap increases pop tht last element
if(minHeap.size()>k) {
minHeap.poll();
}
}
// print remaining element
//HERE IS THE PROBLEM I WANT IT IN " DECREASING ORDER "
// it gives me Increasing order
while(minHeap.size() > 0) {
System.out.print(minHeap.peek()+" ");
minHeap.poll();
}
System.out.println();
}// end of while
}// end of main
}// end of class
GFG类{
公共静态void main(字符串args[])引发IOException{
BufferedReader br=新的BufferedReader(新的InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
而(t-->0){
//取数组和第k个元素iput
字符串n_k[]=br.readLine().split(“”);
//将数组大小存储在k中要查找的n&&k个元素中
int n=Integer.parseInt(n_k[0]);
int k=Integer.parseInt(n_k[1]);
//数组字符串输入
字符串s[]=br.readLine().split(“”);
int d[]=新的int[n];
对于(int i=0;ik){
minHeap.poll();
}
}
//打印剩余元素
//这就是问题所在,我希望它按“降序”排列
//它给了我越来越多的订单
while(minHeap.size()>0){
System.out.print(minHeap.peek()+);
minHeap.poll();
}
System.out.println();
}//结束
}//干管末端
}//下课
输入:
一,
5.2
125787123
输出:
23787只需将代码替换为以下内容即可 在声明优先级队列时,您可以按相反顺序提供比较器 PriorityQueue newHeap=newpriorityqueue(minHeap.size(),Comparator.reverseOrder())强> 在这里,我创建了一个新堆,它将以reverseOrder存储所有现有元素,然后按原样执行所有现有操作 它给出了预期的答案
1
5 2
12 5 787 1 23
O/P
787 23
这对你有帮助
class GFG {
public static void main(String args[]) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
String n_k[] = br.readLine().split(" ");
int n = Integer.parseInt(n_k[0]);
int k = Integer.parseInt(n_k[1]);
String s[] = br.readLine().split(" ");
int d[] = new int[n];
for (int i = 0; i < n; i++) {
d[i] = Integer.parseInt(s[i]);
}
PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>();
for (int i = 0; i < n; i++) {
minHeap.add(d[i]);
// if size of heap increases pop tht last element
if (minHeap.size() > k) {
minHeap.poll();
}
}
PriorityQueue<Integer> newHeap = new PriorityQueue(minHeap.size(), Comparator.reverseOrder());
newHeap.addAll(minHeap);
while (newHeap.size() > 0) {
System.out.print(newHeap.peek() + " ");
newHeap.poll();
}
System.out.println();
}
}
}
GFG类{
公共静态void main(字符串args[])引发IOException{
BufferedReader br=新的BufferedReader(新的InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
而(t-->0){
字符串n_k[]=br.readLine().split(“”);
int n=Integer.parseInt(n_k[0]);
int k=Integer.parseInt(n_k[1]);
字符串s[]=br.readLine().split(“”);
int d[]=新的int[n];
对于(int i=0;ik){
minHeap.poll();
}
}
PriorityQueue newHeap=newpriorityqueue(minHeap.size(),Comparator.reverseOrder());
addAll(minHeap);
while(newHeap.size()>0){
System.out.print(newHeap.peek()+);
newHeap.poll();
}
System.out.println();
}
}
}
您可以使用流
打印结果:
try (Scanner scan = new Scanner(System.in)) {
int totalCases = scan.nextInt();
while (totalCases-- > 0) {
int n = scan.nextInt();
int k = scan.nextInt();
Queue<Integer> minHeap = new PriorityQueue<>(k);
for (int i = 0; i < n; i++) {
if (minHeap.size() == k)
minHeap.remove();
minHeap.add(scan.nextInt());
}
System.out.println(minHeap.stream()
.sorted(Comparator.reverseOrder())
.map(String::valueOf)
.collect(Collectors.joining(" ")));
}
}
try(扫描器扫描=新扫描器(System.in)){
int totalCases=scan.nextInt();
而(总案例-->0){
int n=scan.nextInt();
int k=scan.nextInt();
Queue minHeap=new PriorityQueue(k);
对于(int i=0;i
添加PriorityQueue
不会使空间复杂度达到O(n),如果我错了,请纠正我。谢谢。是的,使用流解决了我在O(1)空间中的问题!非常感谢。