在卡片上找到执行给定操作后剩余的号码
给定一个整数 N ，代表一副牌中的牌数。从 1 到 N 点一副牌，其中 1 是最上面的牌， N 是最下面的牌。你从一副牌中取出最上面的一张牌，并将其插入底部，然后抛出出现在一副牌顶部的下一张牌。你再次做同样的事情，直到剩下一张卡。任务是找到最后剩下的卡的号码。 举例:
Input: N = 4
Output: 1
1 2 3 4
^ ^
Top Bottom
Operation 1: 3 4 1 (1 got shifted to the bottom and 2 got removed)
Operation 2: 1 3 (3 got shifted and 4 got removed)
Operation 3: 1 (3 got removed after shifting 1)
Input: N = 10
Output: 5
进场:
- 首先在队列中插入从 1 到 N 的数字。
- 现在，将前面的元素从队列中出列，并在最后将其入队。
- 最后，弹出前面的元素。
- 打印队列中剩余的最后一个元素。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that will find the
// card number which is remaining
int remainingCard(int n)
{
queue<int> queCards;
// Inserting all the numbers from 1 to n
for (int i = 1; i <= n; i++)
queCards.push(i);
// While there are atleast two
// elements in the queue
while (((int)queCards.size()) >= 2) {
// Push the front element at the back
queCards.push(queCards.front());
// Remove the front element
queCards.pop();
// Remove another element
queCards.pop();
}
// Return the only element
// left in the queue
return queCards.front();
}
// Driver code
int main()
{
int n = 10;
cout << remainingCard(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that will find the
// card number which is remaining
static int remainingCard(int n)
{
Queue<Integer> queCards = new LinkedList<>();
// Inserting all the numbers from 1 to n
for (int i = 1; i <= n; i++)
{
queCards.add(i);
}
// While there are atleast two
// elements in the queue
while (((int) queCards.size()) >= 2)
{
// Push the front element at the back
queCards.add(queCards.peek());
// Remove the front element
queCards.remove();
// Remove another element
queCards.remove();
}
// Return the only element
// left in the queue
return queCards.peek();
}
// Driver code
public static void main(String[] args)
{
int n = 10;
System.out.println(remainingCard(n));
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python3 implementation of the approach
# Function that will find the
# card number which is remaining
def remainingCard(n):
queCards = []
# Inserting all the numbers from 1 to n
for i in range(1, n + 1):
queCards.append(i)
# While there are atleast two
# elements in the queue
while (len(queCards) >= 2):
# Push the front element at the back
queCards.append(queCards[0]);
# Remove the front element
queCards.pop(0);
# Remove another element
queCards.pop(0);
# Return the only element
# left in the queue
return queCards[0]
# Driver code
n = 10
print(remainingCard(n))
# This code is contributed by divyamohan123
C
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function that will find the
// card number which is remaining
static int remainingCard(int n)
{
Queue<int> queCards = new Queue<int>();
// Inserting all the numbers from 1 to n
for (int i = 1; i <= n; i++)
{
queCards.Enqueue(i);
}
// While there are atleast two
// elements in the queue
while (((int) queCards.Count) >= 2)
{
// Push the front element at the back
queCards.Enqueue(queCards.Peek());
// Remove the front element
queCards.Dequeue();
// Remove another element
queCards.Dequeue();
}
// Return the only element
// left in the queue
return queCards.Peek();
}
// Driver code
public static void Main(String[] args)
{
int n = 10;
Console.WriteLine(remainingCard(n));
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// Javascript implementation of the approach
// Function that will find the
// card number which is remaining
function remainingCard(n)
{
queCards = [];
// Inserting all the numbers from 1 to n
for (var i = 1; i <= n; i++)
queCards.push(i);
// While there are atleast two
// elements in the queue
while ((queCards.length) >= 2) {
// Push the front element at the back
queCards.push(queCards[0]);
// Remove the front element
queCards.shift();
// Remove another element
queCards.shift();
}
// Return the only element
// left in the queue
return queCards[0];
}
// Driver code
var n = 10;
document.write( remainingCard(n));
</script>
Output:
5
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