CODE 60: Set Matrix Zeroes

最新推荐文章于 2019-03-31 13:01:40 发布

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最新推荐文章于 2019-03-31 13:01:40 发布

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分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/luzheqi/article/details/12303895

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本文介绍了一种高效的算法，用于在给定的矩阵中如果某个元素为0，则将其所在行和列的所有元素设置为0。该算法采用原地操作的方式，并探讨了如何在不使用额外空间的情况下实现这一功能。

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Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.

click to show follow up.

Follow up:

Did you use extra space?
A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?

	public void setZeroes(int[][] matrix) {
		// Note: The Solution object is instantiated only once and is reused by
		// each test case.
		int zeroH = -1;
		int rows = matrix.length;
		int cols = matrix[0].length;
		if (0 == matrix[0][0]) {
			zeroH = 0;
		}
		for (int i = 1; i < rows; i++) {
			if (0 == matrix[i][0]) {
				matrix[0][0] = 0;
				break;
			}
		}
		for (int i = 1; i < cols; i++) {
			if (0 == matrix[0][i]) {
				zeroH = 0;
				break;
			}
		}
		for (int i = 1; i < rows; i++) {
			for (int j = 1; j < cols; j++) {
				if (0 == matrix[i][j]) {
					matrix[i][0] = 0;
					matrix[0][j] = 0;
				}
			}
		}

		for (int i = 1; i < rows; i++) {
			if (0 == matrix[i][0]) {
				for (int j = 0; j < cols; j++) {
					matrix[i][j] = 0;
				}
			}
		}
		for (int i = 0; i < cols; i++) {
			if (0 == matrix[0][i]) {
				for (int j = 0; j < rows; j++) {
					matrix[j][i] = 0;
				}
			}
		}
		if (0 == zeroH) {
			for (int j = 0; j < cols; j++) {
				matrix[0][j] = 0;
			}
		}
	}