學渣帶你刷Leetcode0059螺旋矩陣 II

題目描述

給定一個正整數 n，生成一個包含 1 到 n2 所有元素，且元素按順時針順序螺旋排列的正方形矩陣。

示例:

輸入: 3
輸出:
[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/spiral-matrix-ii
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白話題目：
 

算法：

 

詳細解釋關注 B站  【C語言全代碼】學渣帶你刷Leetcode 不走丟 https://www.bilibili.com/video/BV1C7411y7gB

C語言完全代碼

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
enum Direction
{
    up = 0,
    right,
    down,
    left,
};

int** generateMatrix(int n, int* returnSize, int** returnColumnSizes){
    if (n <= 0)
    {
        *returnSize = 0;
        *returnColumnSizes = (int*)malloc(1 * sizeof(int));
        (*returnColumnSizes)[*returnSize] = 0;
        return NULL;
    }

    enum Direction direction = 0; // LeetCode平臺定義全局變量會報錯，因此這裏定義枚舉類型
    int uiTop = 0;
    int uiRight = n - 1;
    int uiBottom = n - 1;
    int uiLeft = 0;

    int index = 0;
    int num = 0;

    int** ppRes = (int**)malloc(n * n * sizeof(int*));
    for (index = 0; index <= n - 1; index++)
    {
        ppRes[index] = (int*)malloc(n * sizeof(int));
    }

    *returnSize = n;
    *returnColumnSizes = (int*)malloc(n * sizeof(int));
    for (index = 0; index <= n - 1; index++)
    {
        (*returnColumnSizes)[index] = n;
    }

    while (num < n * n)
    {
        switch (direction)
        {   
            case up:
            {
                for (index = uiLeft; index <= uiRight; index++)
                {
                    ppRes[uiTop][index] = ++num;
                }

                uiTop++;
                break;
            }
            case right:
            {
                for (index = uiTop; index <= uiBottom; index++)
                {
                    ppRes[index][uiRight] = ++num;
                }

                uiRight--;
                break;
            }
            case down:
            {
                for (index = uiRight; index >= uiLeft; index--)
                {
                    ppRes[uiBottom][index] = ++num;
                }
                
                uiBottom--;
                break;
            }
            case left:
            {
                for (index = uiBottom; index >= uiTop; index--)
                {
                    ppRes[index][uiLeft] = ++num;
                }

                uiLeft++;
                break;
            }
            default:
                break;
        }

        direction = (++direction) % 4;
    }

    return ppRes;
}