数组循环移位问题

今天面试遇到的一道题。与参考资料的第一篇文章类似。

另外可以参考STL源码中 rotate() 函数的实现思路。

//TODO 有待填坑

测试几种算法的运行效率

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
public class Solution {

//rotate_right
public static void tripleReverse(int[] nums, int K) {
int length = nums.length;
reverse(nums, 0, length - K - 1);
reverse(nums, length - K, length - 1);
reverse(nums, 0, length - 1);
}

private static void reverse(int[] nums, int start, int end) {
int sum = start + end;
int temp;
for (int i = start; i <= (start + end) / 2; ++i) {
temp = nums[i];
nums[i] = nums[sum - i];
nums[sum - i] = temp;
}
}

//rotate_right
public static void doubleSwap(int[] nums, int K) {
int start = 0;
int end = nums.length;
int middle = nums.length - K;
int i = middle;
int temp;

while (true) {
temp = nums[start];
nums[start] = nums[i];
nums[i] = temp;
++start;
if (++i == end) {
if (start == middle) {
break;
}
i = middle;
} else if (start == middle) {
middle = i;
}
}
}

// rotate left
// if want to rotate right, add [K = nums.length - K] at the beginning
public static void rotate_GCD(int nums[], int K) {
int length = nums.length;
int front, post, temp;
for (int i = 0; i < GCD(length, K); ++i) {
temp = nums[i];
front = i;
while (true) {
post = front + K;
if (post >= length) {
post = post - length;
}
if (post == i) {
break;
}
nums[front] = nums[post];
front = post;
}
nums[front] = temp;
}
}

private static int GCD(int a, int b) {
if (b == 0) {
return a;
} else {
return GCD(b, a % b);
}
}

public static void main(String[] args) {
// 一亿个整数,大概需要三百多兆的内存
int[] array = new int[1_0000_0000];
for (int i = 0; i < 1_0000_0000; ++i) {
array[i] = i;
}

long startTime;
long durationNanoSecond;
double durationMicroSecond;

startTime = System.nanoTime();
tripleReverse(array, 1000000);
durationNanoSecond = System.nanoTime() - startTime;
durationMicroSecond = durationNanoSecond / 1_000;
System.out.println(String.format("[tripleReverse]Time:%f 微秒",
durationMicroSecond));

startTime = System.nanoTime();
doubleSwap(array, 1000000);
durationNanoSecond = System.nanoTime() - startTime;
durationMicroSecond = durationNanoSecond / 1_000;
System.out.println(String.format("[doubleSwap]Time:%f 微秒", durationMicroSecond));

startTime = System.nanoTime();
rotate_GCD(array, 1000000);
durationNanoSecond = System.nanoTime() - startTime;
durationMicroSecond = durationNanoSecond / 1_000;
System.out.println(String.format("[rotate_GCD]Time:%f 微秒", durationMicroSecond));

System.out.print("[result]");
for (int i = 0; i < 10; i++) {
System.out.printf("%d,", array[i]);
}
}
}

参考资料

  1. 数组循环移位,你能想到多少种算法?
  2. STL源码—— rotate算法理解
  3. STL实现细节之rotate()
  4. 《STL源码剖析》学习—6章—_rotate算法分析
  5. STL源码之rotate函数结合图和实例分析
  6. Permutation Algorithms - From Mathematics to Generic Programming (2015)
  7. Program for array rotation