[ 题目描述 ]：
[ 思路 ]：

• 题目要求按顺时针顺序给出m行n列的矩阵的数组
• 按照题目所给的顺序挨个插入答案数组中
• 运行如下
  

int* spiralOrder(int** matrix, int matrixSize, int* matrixColSize, int* returnSize) {
    *returnSize = matrixSize * matrixColSize[0];
    int* ans = (int*)malloc(sizeof(int) * (*returnSize)); 

    int top = 0, bottom = matrixSize - 1;
    int left = 0, right = matrixColSize[0] - 1;
    int index = 0;
    int direction = 0;

    while (top <= bottom && left <= right) {
        if (direction == 0) {
            for (int i = left; i <= right; i++) {
                ans[index++] = matrix[top][i];
            }
            top++;
        } else if (direction == 1) {
            for (int i = top; i <= bottom; i++) {
                ans[index++] = matrix[i][right];
            }
            right--;
        } else if (direction == 2) {
            for (int i = right; i >= left; i--) {
                ans[index++] = matrix[bottom][i];
            }
            bottom--;
        } else if (direction == 3) {
            for (int i = bottom; i >= top; i--) {
                ans[index++] = matrix[i][left];
            }
            left++;
        }
        direction = (direction + 1) % 4;
    }

    return ans;
}

• 时间复杂度O(mn)，空间复杂度O(mn)

[ 官方题解 ]：

• 一、模拟，思路基本同上
• 二、按层模拟，可以将矩阵看成若干层，首先输出最外层的元素，其次输出次外层的元素，直到输出最内层的元素