如何在 O(1) Space 中打印 PriorityQueue 的降序
How to Print decreasing order of the PriorityQueue in O(1) Space
正在解决堆上的问题,我希望使用 PriorityQueue 以降序输出问题。
输入:
1
5 2
12 5 787 1 23
输出:
23 787
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
输入:
1
5 2
12 5 787 1 23
输出:
23787
只需用下面的代码替换您的代码即可。
在声明优先级队列的同时,您可以提供倒序的比较器。
PriorityQueue newHeap = new PriorityQueue(minHeap.size(),Comparator.reverseOrder());
我在这里创建了一个新堆,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。
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();
}
}
}
可以使用Stream
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(" ")));
}
}
正在解决堆上的问题,我希望使用 PriorityQueue 以降序输出问题。
输入:
1
5 2
12 5 787 1 23
输出:
23 787
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
输入:
1
5 2
12 5 787 1 23
输出:
23787
只需用下面的代码替换您的代码即可。
在声明优先级队列的同时,您可以提供倒序的比较器。
PriorityQueue newHeap = new PriorityQueue(minHeap.size(),Comparator.reverseOrder());
我在这里创建了一个新堆,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。
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();
}
}
}
可以使用Stream
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(" ")));
}
}