Java实现完美洗牌算法

1 问题描述
有一个长度为2n的数组{a1,a2,a3,…,an,b1,b2,b3,…,bn}，希望排序后变成{a1,b1,a2,b2,a3,b3,…,an,bn}，请考虑有没有时间复杂度为O(n)而空间复杂度为O(1)的解法。

2 解决方案
2.1位置置换算法
下面算法的时间复杂度为O(n)，空间复杂度为O(n)。

package com.liuzhen.practice;

public class Main {
    //对于数组A第i个位置的元素都最终换到了2*i % len的位置
    public void getLocationReplace(String[] A) {
        int len = A.length;
        String[] temp = new String[len];
        for(int i = 1;i < len;i++)
            temp[(2 * i) % len] = A[i];
        for(int i = 1;i < len;i++)
            A[i] = temp[i];
        for(int i = 1;i < len;i = i + 2) {
            String a1 = A[i];
            A[i] = A[i + 1];
            A[i + 1] = a1;
        }
        return;
    }
    
    
    public static void main(String[] args) {
        Main test = new Main();
        String[] A = {"", "a1", "a2", "a3", "a4", "a5", "b1", "b2", "b3", "b4", "b5"};
        test.getLocationReplace(A);
        for(int i = 1;i < A.length;i++)
            System.out.print(A[i]+" ");
    }
}

运行结果：

a1 b1 a2 b2 a3 b3 a4 b4 a5 b5 

2.2 走环算法
下面算法的时间复杂度为O(n)，空间复杂度为O(1)。

package com.liuzhen.practice;

public class Main1 {
    
    public void CycleLeader(String[] A, int start, int mod) {
        for(int i = start * 2 % mod;i != start;i = i * 2 % mod) {
            String temp = A[i];
            A[i] = A[start];
            A[start] = temp;
        }
        return;
    }
    
    public void Reverse(String[] A, int start, int end) {
        while(start < end) {
            String temp = A[start];
            A[start++] = A[end];
            A[end--] = temp;
        }
        return;
    }
    
    public void RightRotate(String[] A, int start, int m, int n) {
        Reverse(A, start + m + 1, start + n);
        Reverse(A, start + n + 1, start + n + m);
        Reverse(A, start + m + 1, start + n + m);
        return;
    }
    
    public void PerfectShuffle(String[] A) {
        int len = A.length;
        int n = (len - 1) / 2;
        int start = 0;
        while(n > 1) {
            //第1步：找到2*m = 3^k - 1，使得3^k <= len - 1 < 3^(k + 1)
            int k = 0, m = 1;
            for(;(len - 1) / m >= 3;k++, m = m * 3);
            m = m / 2;
        
            //第2步：把数组中的A[m + 1,...,n + m]那部分循环右移m位
            RightRotate(A, start, m, n);
            
            //第3步：对于长度为2*m的数组，刚好有k个圈，每个圈的头部为3^i
            for(int i = 0, t = 1;i < k;i++, t = t * 3)
                CycleLeader(A, t, m * 2 + 1);
            
            //第4步：对数组后面部分A[2m + 1,...,2n]继续递归上面3步
            start = start + m * 2;
            n = n - m;
            
        }
        //n == 1时
        String temp = A[1 + start];
        A[1 + start] = A[2 + start];
        A[2 + start] = temp;
        for(int i = 1;i < len;i = i + 2) {
            String a1 = A[i];
            A[i] = A[i + 1];
            A[i + 1] = a1;
        }
        return;
    }
    
    public static void main(String[] args) {
        Main1 test = new Main1();
        String[] A = {"", "a1", "a2", "a3", "a4", "a5", "b1", "b2", "b3", "b4", "b5"};
        test.PerfectShuffle(A);
        for(int i = 1;i < A.length;i++)
            System.out.print(A[i]+" ");
    }
}

运行结果：

a1 b1 a2 b2 a3 b3 a4 b4 a5 b5