从A1阵列移动,......,一个,B1,...,bn的对A1,B1,..,一个,BN [英] Array movement from a1,..,an,b1,..,bn to a1,b1,..,an,bn
问题描述
今天,我遇到了一个问题,这是真的困惑了我
问题
我有阵只是想:改编[A1,A2,A3 ....一个,B1,B2,B3 ..... BN]
,如何移动数组中的元素将其转移到改编[A1,B1,A2,B2 ......一,BN]
,你应该做的运动-place(空间复杂度应该是恒定
)。
我尽力了思考,并得到一个丑陋的算法就像冒泡排序:
B1向前移动N - 1;
B2向前移动N - 2;
。
。
BN-1向前移动1;
但时间复杂度为O(n 2 ),谁可以给我一个更好的算法? 我找到另外一个更好的方法,就像快速排序:
首先,我们交换元件从(N / 2)到(N)的元素,从B1至B(N / 2);现在我们得到两个独立的子问题,因此,我们可以通过递归解决它。
T(N)= 2T(N / 2)+ O(N)
时间复杂度为O(nlgn)
下面是完整的codeS:
无效swapArray(INT *改编,诠释左,右诠释)
{
INT中期=(左+右)>> 1;
INT TEMP =中端+ 1;
而(左< = MID)
{
掉期(ARR [左++],编曲[临时++]);
}
}
无效arrayMove(INT *改编,诠释磅,诠释乐,INT RB,INT重)
{
如果(LE - 磅< = 0 ||重 - RB< = 0)
返回;
INT中期=(LB +了+ 1)>> 1;
INT LEN =乐 - 中;
如果(RB + LEN> =重)
{
swapArray(ARR,中+ 1,RB);
}
其他
{
swapArray(ARR,中,RB + LEN);
}
arrayMove(ARR,磅,中期 - 1,中,下);
arrayMove(ARR,RB,RB + len个,RB + 1 + len个,重);
}
涉猎和试验/绊倒了一点后,我觉得我开始明白,虽然数学依然对我来说很难。我认为它是这样的:
确定换位的排列周期(这可以在或实际的数据传输之前进行)。式中,来= 2 *从MOD(M * N - 1),其中M = 2,N =数组长度/ 2
,可用于查找指数目的地(排列)。 (Ⅰ减少式为这个问题,因为我们知道,M = 2)的标志访问了索引可以帮助确定下一个周期的开始(从技术上来说,人们可以使用周期计算,而不是一个位集合作为标记,只保留下一周期启动在内存中)。临时变量保存从起始地址的数据,直到周期结束。
总之,这可能意味着两个临时变量的周期计算,并且在就地每个数组元素的一招。
例如:
ARR = 0,1,2,3,4,5,6,7,8,9
目的地:0,2,4,6,8,1,3,5,7,9
开始= 1,TMP = ARR [1]
周期开始
5→1,7-大于5,8-大于7,4-→8,2-→4,tmp-→2
周期结束
没有访问 - 3
开始= 3,TMP =改编[3]
周期开始
6-→3,tmp-→6
周期结束
换位完成。
有问题吗?
随意问及,请访问 http://en.wikipedia.org/wiki/In-place_matrix_transposition
Today, I met a question which is really puzzled me
Question
I have array just like:arr[a1, a2, a3....an, b1, b2, b3.....bn]
, how to move the elements of the array to transfer it into arr[a1, b1, a2, b2......an,bn]
, And you should do the movement in-place(space complexity should be constant
).
I tried my best to think it over and get an ugly algorithm just like bubble-sort:
b1 moves forward by n - 1;
b2 moves forward by n - 2;
.
.
bn-1 moves forward by 1;
But the time complexity is O(n2), who can give me a better algorithm? I find another better method just like quick-Sort:
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)
here are whole codes:
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);
}
After dabbling and experimenting/stumbling a little, I think I'm beginning to understand, although the math is still hard for me. I think it goes something like this:
Determine the permutation cycles of the transposition (this can be done during or before the actual data transfer). The formula, to = 2*from mod (M*N - 1), where M = 2, N = array length / 2
, can be used to find the index destinations (permutation). (I reduced the formula for this question since we know M = 2.) A marker of visited indexes can help determine the start of the next cycle (technically speaking, one could use the cycle calculations rather than a bitset as a marker, keeping only the next cycle-start in memory). A temporary variable holds the data from the start address until the cycle-end.
Altogether, that could mean two temporary variables, the cycle calculations, and one move in-place per array element.
For example:
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.
Any questions?
Feel free to ask and please visit http://en.wikipedia.org/wiki/In-place_matrix_transposition
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