检查队列元素是否成对连续 |组 2
给定一个整数队列。任务是检查队列中的连续元素是否成对连续。
例子：
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Input: 1 2 5 6 9 10
Output: Yes
Input: 2 3 9 11 8 7
Output: No方法 ：
- 取一个变量 n 来存储队列的大小。
- 将元素推送到充当标记的队列。
- 现在，如果队列的大小是奇数，则在 n > 2 时开始弹出一对。此外，在弹出这些对时，检查它们的差异并将 n 减少 2，如果差异不等于 1，则将标志设置为 false。
- 如果队列的大小是偶数，则在 n > 1 时开始弹出一对。此外，在弹出这些对时，检查它们的差异并将 n 减少 2，如果它们的差异不等于 1，则将标志设置为 false。
- 最后，检查标志，如果为真，则表示队列中的元素是成对排序的，否则不是。
下面是上述方法的实现：
C++
// C++ program to check if successive
// pair of numbers in the queue are
// consecutive or not
#include
using namespace std;
// Function to check if elements are
// pairwise consecutive in queue
bool pairWiseConsecutive(queue q)
{
// Fetching size of queue
int n = q.size();
// Pushing extra element to the queue
// which acts as marker
q.push(INT_MIN);
// Result flag
bool result = true;
// If number of elements in the
// queue is odd pop elements while
// its size is greater than 2.
// Else, pop elements while its
// size is greater than 1
if (n % 2 != 0) {
while (n > 2) {
int x = q.front();
q.pop();
q.push(x);
n--;
int y = q.front();
q.pop();
q.push(y);
n--;
if (abs(x - y) != 1) {
result = false;
}
}
// Popping the last element of queue
// and pushing it again.
// Also, popping the extra element
// that we have pushed as marker
q.push(q.front());
q.pop();
q.pop();
}
else {
while (n > 1) {
int x = q.front();
q.pop();
q.push(x);
n--;
int y = q.front();
q.pop();
q.push(y);
n--;
if (abs(x - y) != 1) {
result = false;
}
}
// popping the extra element that
// we have pushed as marker
q.pop();
}
return result;
}
// Driver program
int main()
{
// Pushing elements into the queue
queue q;
q.push(4);
q.push(5);
q.push(-2);
q.push(-3);
q.push(11);
q.push(10);
q.push(5);
q.push(6);
q.push(8);
if (pairWiseConsecutive(q))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
} Java
import java.util.LinkedList;
import java.util.Queue;
// Java program to check if successive
// pair of numbers in the queue are
// consecutive or not
public class GFG {
// Function to check if elements are
// pairwise consecutive in queue
static boolean pairWiseConsecutive(Queue q) {
// Fetching size of queue
int n = q.size();
// Pushing extra element to the queue
// which acts as marker
q.add(Integer.MAX_VALUE);
// Result flag
boolean result = true;
// If number of elements in the
// queue is odd pop elements while
// its size is greater than 2.
// Else, pop elements while its
// size is greater than 1
if (n % 2 != 0) {
while (n > 2) {
int x = q.peek();
q.remove();
q.add(x);
n--;
int y = q.peek();
q.remove();
q.add(y);
n--;
if (Math.abs(x - y) != 1) {
result = false;
}
}
// Popping the last element of queue
// and pushing it again.
// Also, popping the extra element
// that we have pushed as marker
q.add(q.peek());
q.remove();
q.remove();
} else {
while (n > 1) {
int x = q.peek();
q.remove();
q.add(x);
n--;
int y = q.peek();
q.remove();
q.add(y);
n--;
if (Math.abs(x - y) != 1) {
result = false;
}
}
// popping the extra element that
// we have pushed as marker
q.remove();
}
return result;
}
// Driver program
static public void main(String[] args) {
// Pushing elements into the queue
Queue q = new LinkedList<>();
q.add(4);
q.add(5);
q.add(-2);
q.add(-3);
q.add(11);
q.add(10);
q.add(5);
q.add(6);
q.add(8);
if (pairWiseConsecutive(q)) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 program to check if successive
# pair of numbers in the queue are
# consecutive or not
import sys, queue
# Function to check if elements are
# pairwise consecutive in queue
def pairWiseConsecutive(q):
# Fetching size of queue
n = q.qsize()
# Pushing extra element to the
# queue which acts as marker
q.put(-sys.maxsize);
# Result flag
result = bool(True)
# If number of elements in the
# queue is odd pop elements while
# its size is greater than 2.
# Else, pop elements while its
# size is greater than 1
if (n % 2 != 0):
while (n > 2):
x = q.queue[0]
q.get()
q.put(x)
n -= 1
y = q.queue[0]
q.get()
q.put(y)
n -= 1
if (abs(x - y) != 1):
result = bool(False)
# Popping the last element of queue
# and pushing it again.
# Also, popping the extra element
# that we have pushed as marker
q.put(q.queue[0])
q.get()
q.get()
else:
while (n > 1):
x = q.queue[0]
q.get()
q.put(x)
n -= 1
y = q.queue[0]
q.get()
q.put(y)
n -= 1
if (abs(x - y) != 1):
result = bool(False)
# popping the extra element that
# we have pushed as marker
q.get()
return result
# Driver code
# Pushing elements into the queue
q = queue.Queue()
q.put(4)
q.put(5)
q.put(-2)
q.put(-3)
q.put(11)
q.put(10)
q.put(5)
q.put(6)
q.put(8)
if (bool(pairWiseConsecutive(q))):
print("Yes")
else:
print("No")
# This code is contributed by divyeshrabadiya07C#
// C# program to check if successive
// pair of numbers in the queue are
// consecutive or not
using System;
using System.Collections.Generic;
class GFG
{
// Function to check if elements are
// pairwise consecutive in queue
static Boolean pairWiseConsecutive(Queue q)
{
// Fetching size of queue
int n = q.Count;
// Pushing extra element to the queue
// which acts as marker
q.Enqueue(int.MaxValue);
// Result flag
Boolean result = true;
// If number of elements in the
// queue is odd pop elements while
// its size is greater than 2.
// Else, pop elements while its
// size is greater than 1
if (n % 2 != 0)
{
while (n > 2)
{
int x = q.Peek();
q.Dequeue();
q.Enqueue(x);
n--;
int y = q.Peek();
q.Dequeue();
q.Enqueue(y);
n--;
if (Math.Abs(x - y) != 1)
{
result = false;
}
}
// Popping the last element of queue
// and pushing it again.
// Also, popping the extra element
// that we have pushed as marker
q.Enqueue(q.Peek());
q.Dequeue();
q.Dequeue();
} else
{
while (n > 1)
{
int x = q.Peek();
q.Dequeue();
q.Enqueue(x);
n--;
int y = q.Peek();
q.Dequeue();
q.Enqueue(y);
n--;
if (Math.Abs(x - y) != 1)
{
result = false;
}
}
// popping the extra element that
// we have pushed as marker
q.Dequeue();
}
return result;
}
// Driver Code
static public void Main(String[] args)
{
// Pushing elements into the queue
Queue q = new Queue();
q.Enqueue(4);
q.Enqueue(5);
q.Enqueue(-2);
q.Enqueue(-3);
q.Enqueue(11);
q.Enqueue(10);
q.Enqueue(5);
q.Enqueue(6);
q.Enqueue(8);
if (pairWiseConsecutive(q))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by Rajput-Ji 输出：
Yes时间复杂度： O(n)
辅助空间： O(1)