数组从 a1,..,an,b1,..,bn 移动到 a1,b1,..,an,bn

Question

今天遇到一个让我很困惑的问题

问题

我有一个数组，就像：arr[a1, a2, a3....an, b1, b2, b3.....bn]，如何移动数组的元素将它转移到arr[a1, b1, a2, b2......an,bn]，你应该就地移动（space complexity should be constant）。

我尽力想了想，得到了一个像冒泡排序一样丑陋的算法：

b1 moves forward by n - 1;
b2 moves forward by n - 2;
.
.
bn-1 moves forward by 1;

但是时间复杂度是O(n²)，谁能给我一个更好的算法？ 我找到了另一种更好的方法，就像快速排序一样：

First we swap the element from a(n/2) to a(n) with the elements from b1 to b(n/2);now we get two independent sub problems,So we can solve it by recursion.
T(n) = 2T(n/2) + O(n) 
the time complexity is O(nlgn)

这里是完整的代码：

void swapArray(int *arr, int left, int right)
{
    int mid = (left + right) >> 1;
    int temp = mid + 1;
    while(left <= mid)
    {
        swap(arr[left++], arr[temp++]);
    }
}
void arrayMove(int *arr, int lb, int le, int rb, int re)
{
    if(le - lb <= 0 || re - rb <= 0)
        return;
    int mid = (lb + le + 1) >> 1;
    int len = le - mid;
    if(rb + len >= re)
    {
        swapArray(arr, mid + 1, rb);
    }
    else
    {
        swapArray(arr, mid, rb + len);
    }
    arrayMove(arr, lb, mid - 1, mid, le);
    arrayMove(arr, rb, rb + len, rb + 1 + len, re);
}

Answer 1

在涉猎和实验/磕磕绊绊之后，我想我开始明白了，尽管数学对我来说仍然很难。我认为它是这样的：

确定转置的排列周期（这可以在实际数据传输期间或之前完成）。公式to = 2*from mod (M*N - 1), where M = 2, N = array length / 2 可用于查找索引目标（排列）。 （我减少了这个问题的公式，因为我们知道 M = 2。）访问索引的标记可以帮助确定下一个循环的开始（从技术上讲，可以使用循环计算而不是位集作为标记，只保留内存中的下一个循环开始）。一个临时变量保存从起始地址到循环结束的数据。

总而言之，这可能意味着两个临时变量、循环计算和每个数组元素一个原地移动。

例如：

arr          = 0,1,2,3,4,5,6,7,8,9
destinations:  0,2,4,6,8,1,3,5,7,9

start = 1, tmp = arr[1]    

cycle start
5->1, 7->5, 8->7, 4->8, 2->4, tmp->2
cycle end

not visited - 3

start = 3, tmp = arr[3]

cycle start
6->3, tmp->6
cycle end

Transposition complete.

有什么问题吗？
随时询问，请访问http://en.wikipedia.org/wiki/In-place_matrix_transposition

Answer 2

这是一个用 Python 编写的简短代码：

n = 4
a = ['a1', 'a2', 'a3', 'a4', 'b1', 'b2', 'b3', 'b4']
result = a[:n]
for index in range(n):
    result.insert(index*2 + 1, a[n+index])
print result  #  prints ['a1', 'b1', 'a2', 'b2', 'a3', 'b3', 'a4', 'b4']

Answer 3

也许这不是最好的方法，因为它不能直接回答您的问题，但是您为什么不简单地将元素的新索引映射到元素的旧索引？

old_array: [a1, a2, a3....an, b1, b2, b3.....bn]
=>
lookup: [0 n 1 n+1 2 n+2 ... n-1 2n-1]
=>
[a1 b1 a2 b2 ... an bn]

当你想从索引中抓取一个元素时

Get(index)
 return old_array[lookup[index]]
End

Get(2) => old_array[lookup[2]] => old_array[1] => a2 | a2 is at index 2

Answer 4

“从 a1,..,an,b1,..,bn 到 a1,b1,..,an,bn 的数组移动”的工作 Java 代码

private static void myTrans(int[] m) {
    int n = (m.length - 1);

    int i = 1;
    for (int start = 1; start < n; start++) {
        int temp = m[start];
        m[start] = m[n / 2 + i];

        for (int j = (n / 2 + i); j > start; j--) {
            m[j] = m[j - 1];
        }

        start++;
        m[start] = temp;
        printArray(m);
        i++;
    }
}

Answer 5

public void shuffle(char[] arr, int left, int right){
    //center
    int c = (left+right)/2;
    int q = 1 + (left+c)/2;
    for(int k=1,i=q;i<=c;i++,k++){
         //Swap elements
         int temp = arr[i];
         arr[i] = arr[c+k];
         arr[c+k] = temp;
    }
    shuffle(arr,left,c);
    shuffle(arr,c+1,right);
}

这将围绕中心随机播放：

1. a1a2a3a4b1b2b3b4 -> a1a2b1b2a3a4b3b4
2. a1a2b1b2 -> a1b1a2b2
3. a3a4b3b4 -> a3b3a4b4

此解决方案的复杂性将是 nlogn

Answer 6

请检查我尝试的 Java 就地排序解决方案：

时间复杂度：O(N)，其中 N 是数组的长度。 空间复杂度：O(1)

public static void sortSpecialArray(String[] array) {
    sortSpecialArrayUtil(array);
    sortSpecialArrayUtil(array);
}

/**
 * Given an array with ['a1', 'a2', 'a3', ..., 'aN', 'b1', 'b2', 'b3', ...,'bN', 'c1'...'cN']
 * sort the array to be a1, b1, c1...
 *
 * O(N) time complexity (derived from O(N * K) below)
 * O(1) space complexity
 */
private static void sortSpecialArrayUtil(String[] array) {
    // O(1) space complexity requires this function to sort the array in-place
    // this can be done using a pointer that keeps track of where we are
    // in the array. The idea is to swap the values in the array to get the element in the right place.
    int n = stripNumber(array[array.length - 1]);
    int charCount = array.length / n;

    // start the loop at index = 1, up until index = array.length - 2
    // ignore those two elements because they're already in the right place
    int lastIndex = array.length - 2;
    String element;
    for (int i = 1; i <= lastIndex; i++) {
        element = array[i];
        swap(array, i, getCorrectIndex(charCount, element));
    }
}

/**
 * Assume that array's element is in format of something like "a1", where there can only be one 'letter' before the number
 * extract number from the string
 *
 * This is done in time O(K), where K is the length of the String
 */
private static int stripNumber(String str) {
    return Integer.parseInt(str.substring(1, str.length()));
}

private static int getCorrectIndex(int charCount, String element) {
    char c = Character.toLowerCase(element.charAt(0));
    int num = stripNumber(element);

    // the correct index is found by the following formula: index = charDistance + (charCount * (num - 1))
    int charDistance = c - 'a';
    return charDistance + (charCount * (num - 1));
}

private static void swap(String[] array, int fromIndex, int toIndex) {
    String temp = array[toIndex];
    array[toIndex] = array[fromIndex];
    array[fromIndex] = temp;
}

测试：

@Test
public void testSortArray() {
    String[] testCase = new String[] { "a1", "a2", "a3", "a4", "a5", "b1", "b2", "b3", "b4", "b5", "c1", "c2", "c3", "c4", "c5", "d1", "d2", "d3", "d4", "d5" };
    sortSpecialArray(testCase);
    for (String s : testCase) {
        System.out.print(s + " ");
    }
}

Answer 7

假设输入数组为 [a1,a2,a3,b1,b2,b3,c1,c2,c3]，所以根据问题，我们希望输出为 [a1,b1,c1,a2,b2,c2,a3,b3,c3]。让我将数组表示为二维矩阵 每行表示 n 的组大小（在这种情况下 n 为 3）。

Original array
a1 a2 a3
b1 b2 b3
c1 c2 c3

What we Want
a1 b1 c1
a2 b2 c2
a3 b3 c3

你有没有注意到什么。让我告诉你输出数组实际上是输入数组的转置，表示为二维矩阵，可以使用常量空间轻松完成。

• 注意：您需要开发将原始数组分解为二维数组的逻辑。

实现Code