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)。

具体代码如下：

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)。

具体代码如下：

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：