📜  杀死紧邻的邻居后,在 N 人中找到最后 2 名幸存者围成一圈

📅  最后修改于: 2021-09-04 13:11:01             🧑  作者: Mango

给定一个整数N代表N人站在一个圆圈中,任务是找到当一个人以顺时针方向杀死紧邻的邻居时剩下的最后 2 个人。

例子:

朴素的方法:一个简单的方法是保留一个大小为 N 的bool 数组来跟踪一个人是否还活着。

  • 最初,布尔数组对所有人都成立。
  • 保留两个指针,一个指向当前活着的人,第二个存储前一个当前人。
  • 一旦从当前人中找到第二个活着的邻居,将其布尔值更改为 false。
  • 然后再次将当前更新为从前一个活动的下一个。
  • 这个过程会一直持续到最后两个人幸存下来。

时间复杂度: O(N 2 )
辅助空间: O(N)

有效的方法:有效的方法是从数据结构中删除已经死亡的人,这样它就不会被再次遍历。

  • 仅完成一轮后,最多只有 N/2 人。
  • 然后在下一轮中,它将留下 N/4 个人,依此类推,直到活着的人数变为 2。

下面是上述方法的实现:

C++
// C++ implementation of the approach
 
#include 
using namespace std;
 
// Node for a Linked List
struct Node {
    int val;
    struct Node* next;
 
    Node(int _val)
    {
        val = _val;
        next = NULL;
    }
};
 
// Function to find the last 2 survivors
void getLastTwoPerson(int n)
{
    // Total is the count
    // of alive people
    int total = n;
    struct Node* head = new Node(1);
    struct Node* temp = head;
 
    // Initiating the list of n people
    for (int i = 2; i <= n; i++) {
        temp->next = new Node(i);
        temp = temp->next;
    }
 
    temp->next = head;
    temp = head;
 
    struct Node* del;
 
    // Total != 2 is terminating
    // condition because
    // at last only two-person
    // will remain alive
    while (total != 2) {
 
        // Del represent next person
        // to be deleted or killed
        del = temp->next->next;
        temp->next->next
            = temp->next->next->next;
        temp = temp->next;
        free(del);
        total -= 1;
    }
 
    // Last two person to
    // survive (in any order)
    cout << temp->val << " "
         << temp->next->val;
}
 
// Driver code
int main()
{
    int n = 2;
 
    getLastTwoPerson(n);
 
    return 0;
}


Java
// Java implementation of the approach
class GFG{
 
// Node for a Linked List
static class Node
{
    int val;
    Node next;
 
    Node(int _val)
    {
        val = _val;
        next = null;
    }
};
 
// Function to find the last 2 survivors
static void getLastTwoPerson(int n)
{
     
    // Total is the count
    // of alive people
    int total = n;
    Node head = new Node(1);
    Node temp = head;
 
    // Initiating the list of n people
    for(int i = 2; i <= n; i++)
    {
        temp.next = new Node(i);
        temp = temp.next;
    }
 
    temp.next = head;
    temp = head;
 
    Node del;
 
    // Total != 2 is terminating
    // condition because
    // at last only two-person
    // will remain alive
    while (total != 2)
    {
 
        // Del represent next person
        // to be deleted or killed
        del = temp.next.next;
        temp.next.next = temp.next.next.next;
        temp = temp.next;
        del = null;
         
        System.gc();
        total -= 1;
    }
 
    // Last two person to
    // survive (in any order)
    System.out.print(temp.val + " " +
                     temp.next.val);
}
 
// Driver code
public static void main(String[] args)
{
    int n = 2;
 
    getLastTwoPerson(n);
}
}
 
// This code is contributed by Rajput-Ji


Python3
# Python3 implementation of the approach
 
# Node for a Linked List
class newNode:
     
    def __init__(self, val):
         
        self.val = val
        self.next = None
 
# Function to find the last 2 survivors
def getLastTwoPerson(n):
     
    # Total is the count
    # of alive people
    total = n
    head = newNode(1)
    temp = head
 
    # Initiating the list of n people
    for i in range(2, n + 1, 1):
        temp.next = newNode(i)
        temp = temp.next
 
    temp.next = head
    temp = head
 
    de = None
 
    # Total != 2 is terminating
    # condition because
    # at last only two-person
    # will remain alive
    while (total != 2):
         
        # de represent next person
        # to be deeted or killed
        de = temp.next.next
        temp.next.next = temp.next.next.next
        temp = temp.next
        del de
        total -= 1
 
    # Last two person to
    # survive (in any order)
    print(temp.val, temp.next.val)
 
# Driver code
if __name__ == '__main__':
     
    n = 2
     
    getLastTwoPerson(n)
 
# This code is contributed by SURENDRA_GANGWAR


C#
// C# implementation of the approach
using System;
 
class GFG{
 
// Node for a Linked List
class Node
{
    public int val;
    public Node next;
 
    public Node(int _val)
    {
        val = _val;
        next = null;
    }
};
 
// Function to find the last 2 survivors
static void getLastTwoPerson(int n)
{
     
    // Total is the count
    // of alive people
    int total = n;
    Node head = new Node(1);
    Node temp = head;
 
    // Initiating the list of n people
    for(int i = 2; i <= n; i++)
    {
        temp.next = new Node(i);
        temp = temp.next;
    }
 
    temp.next = head;
    temp = head;
 
    Node del;
 
    // Total != 2 is terminating
    // condition because
    // at last only two-person
    // will remain alive
    while (total != 2)
    {
 
        // Del represent next person
        // to be deleted or killed
        del = temp.next.next;
        temp.next.next = temp.next.next.next;
        temp = temp.next;
        del = null;
         
        total -= 1;
    }
 
    // Last two person to
    // survive (in any order)
    Console.Write(temp.val + " " +
                  temp.next.val);
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 2;
 
    getLastTwoPerson(n);
}
}
 
// This code is contributed by Amit Katiyar


输出:
1 2

时间复杂度: O(N*log N)
辅助空间: O(N)

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