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55. Set Matrix Zeroes

2024-07-21 00:20:42   222人阅读

Set Matrix Zeroes

(Link: https://oj.leetcode.com/problems/set-matrix-zeroes/)

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?

思路： 找到第一个 0 元素，记下其行和列。然后对其他 0 元素，分别将其投影在记下的行和列上（做标记）。遍历之后，对于所有行中的标记，将其所在列都置为 0； 对于所有列中标记，将其所在行都置为 0. （最后置标记的行和列为 0. 可含在上述步骤） 时间: O(n2), 空间 : O(1)

class Solution {public:	void setZeroes(vector<vector<int> > &matrix) {		if(!matrix.size() || !matrix[0].size()) return;		const int INF = 1001;		int row = matrix.size(), col = matrix[0].size();		int u = -1, v = -1;		for(int r = 0; r < row; ++r) 			for(int c = 0; c < col; ++c) {				if(matrix[r][c] == 0) {					if(u == -1) {  u = r, v = c; continue; }					matrix[u][c] = INF;					matrix[r][v] = INF;				}			}			if(u == -1) return;			if(matrix[u][v] == INF) matrix[u][v] = 0;			for(int c = 0; c < col; ++c) {				if(matrix[u][c] == INF) {					for(int i = 0; i < row; ++i)  matrix[i][c] = 0;				} else matrix[u][c] = 0;			}			for(int r = 0; r < row; ++r) {				if(matrix[r][v] == INF) {					for(int j = 0; j < col; ++j) matrix[r][j] = 0;				} else matrix[r][v] = 0;			}	}};

 

 

55. Set Matrix Zeroes


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