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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
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