Question

我有一个数组

a[]= {34,45,65,55,67}

我需要C或TCL代码来构建一个新数组，每个元素重复“N”次。

例如，当n = 2时，结果数组应为

b[]= {34,34,45,45,65,65,55,55,67,67}

同样，当n = 3时，数组应为

b[]={34,34,34,45,45,45,65,65,65,55,55,55,67,67,67}

我该怎么做？

Answer 1

// Input parameters
int i,j;
int a[] = {34,45,65,55,67};
int aSize = 5;
int repeat = 10;

// Create a new array with a dynamic size.
// This array must be freed after to avoid memory leaks
int b* = (int*) malloc(sizeof(int) * aSize * repeat);

for (i = 0; i < aSize; ++i) // for all elements in a
    for (j = 0; j < repeat; ++j) // repeat them "repeat" times
        b[i * repeat + j] = a[i];  // i * repeat + j is the current element in b

// do something with b here

// release memory
free(b);

Answer 2

这是一个TCL解决方案：

% package require struct
2.1
% proc dup {items count} {
     return [join [struct::list mapfor x $items {struct::list repeat $count $x}]]
}
% dup {1 2 3} 3
1 1 1 2 2 2 3 3 3

构建数组并添加元素'N'次

2 个答案: