Josephus Circle 使用 STL 列表实现
有n人围成一圈等待被处决。计数从圆圈中的某个点开始，并以固定方向绕圆圈进行。每一步都会跳过一定数量的人，然后执行下一个人。消除围绕圆圈进行（随着被处决的人被移除，圆圈越来越小），直到只剩下最后一个人，他获得了自由。给定总人数 n 和数字 k，表示跳过 k-1 个人，并在圈内杀死第 k 个人。任务是选择初始圈中的位置，以便您是最后一个剩余的并生存。 （基于 0 的索引）。
例子 ：
Input : Length of circle : n = 4
Count to choose next : k = 2
Output : 0
Input : n = 5
k = 3
Output : 3我们已经讨论了这个问题的不同解决方案（这里，这里和这里）。在这篇文章中，讨论了使用列表容器的基于 C++ STL 的解决方案，该解决方案使用循环列表的思想。
C++
// CPP program to find last man standing
#include
using namespace std;
int josephusCircle(int n, int k){
listl; //creates a doubly linked list using stl container//
for(int i=0;i1){
for(int i=1;i Java
// Java Code to find the last man Standing
public class GFG {
// Node class to store data
static class Node
{
public int data ;
public Node next;
public Node( int data )
{
this.data = data;
}
}
/* Function to find the only person left
after one in every m-th node is killed
in a circle of n nodes */
static void getJosephusPosition(int m, int n)
{
// Create a circular linked list of
// size N.
Node head = new Node(1);
Node prev = head;
for(int i = 2; i <= n; i++)
{
prev.next = new Node(i);
prev = prev.next;
}
// Connect last node to first
prev.next = head;
/* while only one node is left in the
linked list*/
Node ptr1 = head, ptr2 = head;
while(ptr1.next != ptr1)
{
// Find m-th node
int count = 1;
while(count != m)
{
ptr2 = ptr1;
ptr1 = ptr1.next;
count++;
}
/* Remove the m-th node */
ptr2.next = ptr1.next;
ptr1 = ptr2.next;
}
System.out.println ("Last person left standing " +
"(Josephus Position) is " + ptr1.data);
}
/* Driver program to test above functions */
public static void main(String args[])
{
int n = 14, m = 2;
getJosephusPosition(m, n);
}
}Python3
# Python3 program to find last man standing
# /* structure for a node in circular
# linked list */
class Node:
def __init__(self, x):
self.data = x
self.next = None
# /* Function to find the only person left
# after one in every m-th node is killed
# in a circle of n nodes */
def getJosephusPosition(m, n):
# Create a circular linked list of
# size N.
head = Node(1)
prev = head
for i in range(2, n + 1):
prev.next = Node(i)
prev = prev.next
prev.next = head # Connect last
#node to first
#/* while only one node is left in the
#linked list*/
ptr1 = head
ptr2 = head
while (ptr1.next != ptr1):
# Find m-th node
count = 1
while (count != m):
ptr2 = ptr1
ptr1 = ptr1.next
count += 1
# /* Remove the m-th node */
ptr2.next = ptr1.next
# free(ptr1)
ptr1 = ptr2.next
print("Last person left standing (Josephus Position) is ", ptr1.data)
# /* Driver program to test above functions */
if __name__ == '__main__':
n = 14
m = 2
getJosephusPosition(m, n)
# This code is contributed by mohit kumar 29C#
// C# Code to find the last man Standing
using System;
public class GFG {
// Node class to store data
class Node
{
public int data ;
public Node next;
public Node( int data )
{
this.data = data;
}
}
/* Function to find the only person left
after one in every m-th node is killed
in a circle of n nodes */
static void getJosephusPosition(int m, int n)
{
// Create a circular linked list of
// size N.
Node head = new Node(1);
Node prev = head;
for(int i = 2; i <= n; i++)
{
prev.next = new Node(i);
prev = prev.next;
}
// Connect last node to first
prev.next = head;
/* while only one node is left in the
linked list*/
Node ptr1 = head, ptr2 = head;
while(ptr1.next != ptr1)
{
// Find m-th node
int count = 1;
while(count != m)
{
ptr2 = ptr1;
ptr1 = ptr1.next;
count++;
}
/* Remove the m-th node */
ptr2.next = ptr1.next;
ptr1 = ptr2.next;
}
Console.WriteLine ("Last person left standing " +
"(Josephus Position) is " + ptr1.data);
}
/* Driver program to test above functions */
static public void Main(String []args)
{
int n = 14, m = 2;
getJosephusPosition(m, n);
}
}
//contributed by Arnab KunduJavascript
输出
12输出 ：
Last person left standing (Josephus Position) is 12 ( 0 based indexing )如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。