如何将一个长度为 n 的数组分成 k 个大小大致相等的子数组(在 C 中)?
How to divide an array of n length into k sub arrays that have approximately equal sizes (in C)?
我有一个大小为n的数组,想分成k个子数组,每个数组的大小必须大致相同。我想了一会儿,知道你必须使用两个 for 循环,但我很难实现这些 for 循环。
我尝试过的:
//Lets call the original integer array with size n: arr
// n is the size of arr
// k is the number of subarrays wanted
int size_of_subArray = n/k;
int left_over = n%k; // When n is not divisible by k
int list_of_subArrays[k][size_of_subArray + 1];
//Lets call the original integer array with size n: arr
for(int i = 0; i < k; i++){
for(int j = 0; j < size_of_subArray; j++){
list_of_subArrays[i][j] = arr[j];
}
}
我正在努力在 forloop 中获取正确的索引。
有什么想法吗?
我已经重构了你的代码并对其进行了注释。
要点是:
- 计算子数组大小时,必须向上取整
arr
的索引需要从0继续递增(即不是重置为0)
以下应该有效,但我没有测试它[请原谅无偿样式清理]:
// Lets call the original integer array with size n: arr
// n is the size of arr
// k is the number of subarrays wanted
// round up the size of the subarray
int subsize = (n + (k - 1)) / k;
int list_of_subArrays[k][subsize];
int arridx = 0;
int subno = 0;
// process all elements in original array
while (1) {
// get number of remaining elements to process in arr
int remain = n - arridx;
// stop when done
if (remain <= 0)
break;
// clip remaining count to amount per sub-array
if (remain > subsize)
remain = subsize;
// fill next sub-array
for (int subidx = 0; subidx < remain; ++subidx, ++arridx)
list_of_subArrays[subno][subidx] = arr[arridx];
// advance to next sub-array
++subno;
}
更新:
Yes this divides the arrays into n subarrays, but it doesn't divide it evenly. Say there was an array of size 10, and wanted to divide it into 9 subarrays. Then 8 subarrays will have 1 of original array's element, but one subarray will need to have 2 elements.
您的原始代码有一些错误[已在上面的示例中修复]。即使我是为自己做这件事,上面的内容也是开始工作的第一步。
在你原来的问题中,你确实说:"and each array must have approximately the same size"。但是,这里有列表子数组的物理大小[仍然是向上舍入的值]。
但是,我可能会说 "evenly distributed" 之类的话来进一步阐明您的意图。也就是说,您希望最后一个 sub-array/bucket 到 而不是 是 "short" [大幅度]。
鉴于此,代码开始时有些相同,但需要更复杂一些。这仍然有点粗糙,可能会进一步优化:
#include <stdio.h>
#ifdef DEBUG
#define dbgprt(_fmt...) printf(_fmt)
#else
#define dbgprt(_fmt...) /**/
#endif
int arr[5000];
// Lets call the original integer array with size n: arr
// n is the size of arr
// k is the number of subarrays wanted
void
fnc2(int n,int k)
{
// round up the size of the subarray
int subsize = (n + (k - 1)) / k;
int list_of_subArrays[k][subsize];
dbgprt("n=%d k=%d subsize=%d\n",n,k,subsize);
int arridx = 0;
for (int subno = 0; subno < k; ++subno) {
// get remaining number of sub-arrays
int remsub = k - subno;
// get remaining number of elements
int remain = n - arridx;
// get maximum bucket size
int curcnt = subsize;
// get projected remaining size for using this bucket size
int curtot = remsub * curcnt;
// if we're too low, up it
if (curtot < remain)
++curcnt;
// if we're too high, lower it
if (curtot > remain)
--curcnt;
// each bucket must have at least one
if (curcnt < 1)
curcnt = 1;
// each bucket can have no more than the maximum
if (curcnt > subsize)
curcnt = subsize;
// last bucket is the remainder
if (curcnt > remain)
curcnt = remain;
dbgprt(" list[%d][%d] --> arr[%d] remain=%d\n",
subno,curcnt,arridx,remain);
// fill next sub-array
for (int subidx = 0; subidx < curcnt; ++subidx, ++arridx)
list_of_subArrays[subno][subidx] = arr[arridx];
}
dbgprt("\n");
}
我有一个大小为n的数组,想分成k个子数组,每个数组的大小必须大致相同。我想了一会儿,知道你必须使用两个 for 循环,但我很难实现这些 for 循环。
我尝试过的:
//Lets call the original integer array with size n: arr
// n is the size of arr
// k is the number of subarrays wanted
int size_of_subArray = n/k;
int left_over = n%k; // When n is not divisible by k
int list_of_subArrays[k][size_of_subArray + 1];
//Lets call the original integer array with size n: arr
for(int i = 0; i < k; i++){
for(int j = 0; j < size_of_subArray; j++){
list_of_subArrays[i][j] = arr[j];
}
}
我正在努力在 forloop 中获取正确的索引。
有什么想法吗?
我已经重构了你的代码并对其进行了注释。
要点是:
- 计算子数组大小时,必须向上取整
arr
的索引需要从0继续递增(即不是重置为0)
以下应该有效,但我没有测试它[请原谅无偿样式清理]:
// Lets call the original integer array with size n: arr
// n is the size of arr
// k is the number of subarrays wanted
// round up the size of the subarray
int subsize = (n + (k - 1)) / k;
int list_of_subArrays[k][subsize];
int arridx = 0;
int subno = 0;
// process all elements in original array
while (1) {
// get number of remaining elements to process in arr
int remain = n - arridx;
// stop when done
if (remain <= 0)
break;
// clip remaining count to amount per sub-array
if (remain > subsize)
remain = subsize;
// fill next sub-array
for (int subidx = 0; subidx < remain; ++subidx, ++arridx)
list_of_subArrays[subno][subidx] = arr[arridx];
// advance to next sub-array
++subno;
}
更新:
Yes this divides the arrays into n subarrays, but it doesn't divide it evenly. Say there was an array of size 10, and wanted to divide it into 9 subarrays. Then 8 subarrays will have 1 of original array's element, but one subarray will need to have 2 elements.
您的原始代码有一些错误[已在上面的示例中修复]。即使我是为自己做这件事,上面的内容也是开始工作的第一步。
在你原来的问题中,你确实说:"and each array must have approximately the same size"。但是,这里有列表子数组的物理大小[仍然是向上舍入的值]。
但是,我可能会说 "evenly distributed" 之类的话来进一步阐明您的意图。也就是说,您希望最后一个 sub-array/bucket 到 而不是 是 "short" [大幅度]。
鉴于此,代码开始时有些相同,但需要更复杂一些。这仍然有点粗糙,可能会进一步优化:
#include <stdio.h>
#ifdef DEBUG
#define dbgprt(_fmt...) printf(_fmt)
#else
#define dbgprt(_fmt...) /**/
#endif
int arr[5000];
// Lets call the original integer array with size n: arr
// n is the size of arr
// k is the number of subarrays wanted
void
fnc2(int n,int k)
{
// round up the size of the subarray
int subsize = (n + (k - 1)) / k;
int list_of_subArrays[k][subsize];
dbgprt("n=%d k=%d subsize=%d\n",n,k,subsize);
int arridx = 0;
for (int subno = 0; subno < k; ++subno) {
// get remaining number of sub-arrays
int remsub = k - subno;
// get remaining number of elements
int remain = n - arridx;
// get maximum bucket size
int curcnt = subsize;
// get projected remaining size for using this bucket size
int curtot = remsub * curcnt;
// if we're too low, up it
if (curtot < remain)
++curcnt;
// if we're too high, lower it
if (curtot > remain)
--curcnt;
// each bucket must have at least one
if (curcnt < 1)
curcnt = 1;
// each bucket can have no more than the maximum
if (curcnt > subsize)
curcnt = subsize;
// last bucket is the remainder
if (curcnt > remain)
curcnt = remain;
dbgprt(" list[%d][%d] --> arr[%d] remain=%d\n",
subno,curcnt,arridx,remain);
// fill next sub-array
for (int subidx = 0; subidx < curcnt; ++subidx, ++arridx)
list_of_subArrays[subno][subidx] = arr[arridx];
}
dbgprt("\n");
}