c 获取数组的全排列

根据给定的int型数组，给出全排列

比如: [1,2] => [[1,2],[2,1] [1,2,3] =>[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

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


int *concat(int* list, int length, int x) {
	int *result = (int*)malloc((length + 1) * sizeof(int));
	for (int i = length; i > 0; i--) {
		result[i] = list[i - 1];
	}
	result[0] = x;
	return result;
}

int getFactorial(int n) {
	int total = 1;
	for (int i = 1; i <= n; i++) {
		total *= i;
	}
	return total;
}

int** getPermutation(int * list ,int n) {
	int** result;
	int total = getFactorial(n);
	if (n == 2) {
		result = (int**)malloc(total * sizeof(int*));
		for (int i = 0; i < total; i++) {
			result[i] = (int*)malloc(n * sizeof(int));
		}
		result[0][0] = list[0];
		result[0][1] = list[1];
		result[1][0] = list[1];
		result[1][1] = list[0];
		return result;
	}else{
		result = (int**)malloc(total * sizeof(int*));
		for (int i = 0; i < n; i++) {
			int* tempList = (int*)malloc((n - 1) * sizeof(int));
			int curIdx = 0;
			for (int j = 0; j < n; j++) {
				if (i == j) continue;
				tempList[curIdx] = list[j];
				curIdx++;
			}
			int** tempResult = getPermutation(tempList, n - 1);
			int tempResultLength = getFactorial(n - 1);
			for (int k = 0; k < tempResultLength; k++) {
				result[tempResultLength * i + k] = concat(tempResult[k], n - 1, list[i]);	
			}
			free(tempList);
			free(tempResult);
		}
		return result;
	}
}

int main() {
	int n =4;
	int* p = (int*)malloc(n * sizeof(int));
	for (int i = 0; i < n; i++) {
		p[i] = i + 1;
	}
	
	int **result = getPermutation(p,n);

	for (int i = 0; i < getFactorial(n); i++) {
		for (int j = 0; j < n; j++) {
			printf("%d,", result[i][j]);
		}
		printf("\n");
	}

	free(p);
	free(result);

	return 0;
}