请你分别写出对应的伪代码和c语言完整代码

伪代码：

function solve_n_queens(n):
    function is_valid(board, row, col):
        for i from 0 to row - 1:
            if board[i] == col or abs(board[i] - col) == row - i:
                return False
        return True

    function backtrack(board, row):
        if row == n:
            result.append(board[:])
        else:
            for col from 0 to n - 1:
                if is_valid(board, row, col):
                    board[row] = col
                    backtrack(board, row + 1)
                    board[row] = -1

    result = []
    board = [-1] * n
    backtrack(board, 0)
    return result

C语言完整代码：

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

bool is_valid(int *board, int row, int col, int n) {
    for (int i = 0; i < row; i++) {
        if (board[i] == col || abs(board[i] - col) == row - i) {
            return false;
        }
    }
    return true;
}

void backtrack(int *board, int row, int n, int **result, int *count) {
    if (row == n) {
        result[*count] = (int *)malloc(n * sizeof(int));
        for (int i = 0; i < n; i++) {
            result[*count][i] = board[i];
        }
        (*count)++;
    } else {
        for (int col = 0; col < n; col++) {
            if (is_valid(board, row, col, n)) {
                board[row] = col;
                backtrack(board, row + 1, n, result, count);
                board[row] = -1;
            }
        }
    }
}

int **solve_n_queens(int n, int *returnSize, int **returnColumnSizes) {
    int **result = (int **)malloc(100 * sizeof(int *));
    int count = 0;
    int *board = (int *)malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        board[i] = -1;
    }
    backtrack(board, 0, n, result, &count);
    *returnSize = count;
    *returnColumnSizes = (int *)malloc(count * sizeof(int));
    for (int i = 0; i < count; i++) {
        (*returnColumnSizes)[i] = n;
    }
    return result;
}

int main() {
    int n = 4;
    int returnSize;
    int *returnColumnSizes;
    int **result = solve_n_queens(n, &returnSize, &returnColumnSizes);
    printf("[\n");
    for (int i = 0; i < returnSize; i++) {
        printf("  [");
        for (int j = 0; j < n; j++) {
            printf("%d", result[i][j]);
            if (j < n - 1) {
                printf(", ");
            }
        }
        printf("]");
        if (i < returnSize - 1) {
            printf(",");
        }
        printf("\n");
    }
    printf("]\n");
    return 0;
}

输出为：

[
  [1, 3, 0, 2],
  [2, 0, 3, 1]
]