[Daily Practice] - Josephus problem

Problem Description

0,1, n-1 which are arranged in a circle numbers n, starting from the number 0, delete the m-th digit each time from inside the circle. Find the circle where the last remaining digit.
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For example, 0,1,2,3,4 five numbers in a circle, each deletion of three numbers starting from the number 0, the first four digits of the deleted sequence is 2,0,4,1, so the final the remaining number is 3.

A Solution

More intuitive solution

public int lastRemaining(int n, int m) {
        LinkedList<Integer> list = new LinkedList<>();
        // 全部加进去
        for (int i = 0; i < n; i++) {
            list.add(i);
        }
        // 只要链表长度不为 1，就不断循环
        while (list.size() != 1) {
            for (int i = 0; i < m; i++) {
            	// 取出第一个元素并删除
                Integer pre = list.pollFirst();
                if (i != m - 1) {
                	// 如果不是第m个，就加到末尾
                    list.add(pre);
                }
            }
        }
        return list.pollFirst();
}

Solution two

First, we set the n m-bit values continuously deleted, the resulting value of f(n,m)
the first digital deleted is (m-1)%ndenoted x
then the remaining n-1 becomes a digital (assumed here that n is greater than 2)

0	1	…	x-1	x+1	…	n-1

Because of the need from a recount after the number is deleted, we first into X + 1, the new numbers are arranged in the order

The original array	Corresponding to the ordinal position
x+1	0
x+2	1
…	…
n-1	n-x-2
0	n-x-1
…	…
x-1	n-2

Why position 0 corresponds to nx-1?
The first step when we deleted a number that can be understood as this list is divided into two parts, the
length of the first part of x (because it is starting from 0),
then the length of the second part of it is n-x-1
now the first part of the mosaic behind the second portion,
then the first number of the first division, i.e. 0, becomes a n-x-1+1digital, i.e. numbers nx, because it is starting from zero, so that the corresponding n-x-1
key points:
Suppose the sequence corresponding to position a, the original value of B
then the relationship sequential positions a and B values, we can draw the formula:

B = (A+x+1) mod n

Can be obtained from the initial definition:
n-th digit. 1-m-th continuously remove the last value obtained f(n-1,m)
from the above equation we can be pushed back to its inverse value of （f(n-1,m)+x+1)%n
the value defined at the beginning is that we f(n-1,m)
will be x=(m-1)%nsubstituted into simplified available :

f(n,m)=（f(n−1,m)+m) mod n，且f(1,m) = 0

Code

 public static int lastRemaining(int n, int m) {
        return n == 1 ? 0 : (lastRemaining(n-1,m) + m) % n;
    }

or

public int lastRemaining(int n, int m) {
    int flag = 0;   
    for (int i = 2; i <= n; i++) {
        flag = (flag + m) % i;
    }
    return flag;
}