054 - Spiral Matrix

最新推荐文章于 2021-09-18 15:12:15 发布

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本文介绍了一种螺旋遍历矩阵的算法实现，该算法能够按螺旋顺序返回给定矩阵的所有元素。例如，对于一个3x3的矩阵[[1,2,3],[4,5,6],[7,8,9]]，输出结果应为[1,2,3,6,9,8,7,4,5]。文章提供了详细的C语言代码实现。

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Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

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

You should return [1,2,3,6,9,8,7,4,5].

int* spiralOrder(int** matrix, int matrixRowSize, int matrixColSize) 
{
	int *ret = (int *)malloc(sizeof(int) * matrixRowSize * matrixColSize);
	int k = 0, i;
	int over = matrixRowSize > matrixColSize? matrixColSize : matrixRowSize;
	over -= 1;
	int low = 0;

	int north = 0;
	int south = matrixRowSize - 1;
	int east = matrixColSize - 1;
	int west = 0;
	while (low <= over) {
		for (i = west; i <= east; i++)
			ret[k++] = matrix[north][i];
		for (i = north + 1; i <= south - 1; i++)
			ret[k++] = matrix[i][east];
		for (i = east; north != south && i >= west; i--)
			ret[k++] = matrix[south][i];
		for (i = south - 1; west != east && i >= north + 1; i--)
			ret[k++] = matrix[i][west];
		north++,south--;
		west++,east--;
		low++, over--;
	}

	return ret;
}