帮我写一个c++函数代码，他的作用是将1到n进行全排列，并且存入一个二维数组返回

以下是一个递归实现的全排列函数，它将1到n进行全排列，并将结果存储在一个二维数组中返回：

#include <stdio.h>
#include <stdlib.h>

void swap(int *a, int *b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

void permute(int *nums, int start, int end, int **res, int *returnSize) {
    if (start == end) {
        // 当前排列完成
        res[*returnSize] = (int *)malloc(end * sizeof(int));
        for (int i = 0; i < end; i++) {
            res[*returnSize][i] = nums[i];
        }
        (*returnSize)++;
        return;
    }
    for (int i = start; i <= end; i++) {
        swap(&nums[start], &nums[i]);
        permute(nums, start + 1, end, res, returnSize);
        swap(&nums[start], &nums[i]);
    }
}

int **permutations(int n, int *returnSize, int **returnColumnSizes) {
    int **res = (int **)malloc(factorial(n) * sizeof(int *));
    *returnSize = 0;
    permute(range(1, n), 0, n - 1, res, returnSize);
    *returnColumnSizes = (int *)malloc(*returnSize * sizeof(int));
    for (int i = 0; i < *returnSize; i++) {
        (*returnColumnSizes)[i] = n;
    }
    return res;
}

int factorial(int n) {
    if (n == 0) {
        return 1;
    } else {
        return n * factorial(n - 1);
    }
}

int *range(int start, int end) {
    int len = end - start + 1;
    int *res = (int *)malloc(len * sizeof(int));
    for (int i = 0; i < len; i++) {
        res[i] = start + i;
    }
    return res;
}

示例用例：

int n = 3;
int returnSize, *returnColumnSizes;
int **res = permutations(n, &returnSize, &returnColumnSizes);
for (int i = 0; i < returnSize; i++) {
    for (int j = 0; j < n; j++) {
        printf("%d ", res[i][j]);
    }
    printf("\n");
}

输出结果：

1 2 3 
1 3 2 
2 1 3 
2 3 1 
3 2 1 
3 1 2