LeetCode之Set Matrix Zeroes

最新推荐文章于 2024-08-16 09:55:14 发布

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本文介绍了一种高效的矩阵零填充算法，该算法可以在不使用额外空间的情况下，将矩阵中元素为0的整行和整列设置为0。文章详细解释了如何利用矩阵的第一行和第一列来记录哪些行和列需要被置零，并通过两个布尔变量来判断是否需要将第一行和第一列也置零。

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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.

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?

LeetCode Discuss

“The basic idea is to use the first row and column to store the information.

Then we need to know whether or not to set the first row and column to zeroes.

So, we use two boolean to store that information.”

class Solution {
public:
    void setZeroes(vector<vector<int> > &matrix) {
        int rownum = matrix.size();
		if (rownum == 0) return;
		int colnum = matrix[0].size();
		bool hasZeroFirstRow = false, hasZeroFirstColumn = false;

		for (int i = 0; i < rownum; i++)
			if (matrix[i][0] == 0){
				hasZeroFirstColumn = true;
				break;
			}

		for (int i = 0; i < colnum; i++)
			if (matrix[0][i] == 0){
				hasZeroFirstRow = true;
				break;
			}

		for (int i = 1; i < rownum; i++)
			for (int j = 1; j < colnum; j++){
				if (matrix[i][j] == 0){
					matrix[i][0] = 0;
					matrix[0][j] = 0;
				}
			}

		for (int i = 1; i < rownum; i++)
			for (int j = 1; j < colnum; j++){
				if (matrix[0][j] == 0 || matrix[i][0]==0)
					matrix[i][j] = 0;
				
			}
		
		if (hasZeroFirstRow) {
			for (int j = 0; j < colnum; ++j) 
				matrix[0][j] = 0;
		}

		if (hasZeroFirstColumn) {
			for (int i = 0; i < rownum; ++i)
				matrix[i][0] = 0;
		}
    }
};