leetcode--Set Matrix Zeroes

最新推荐文章于 2023-02-25 23:03:36 发布

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最新推荐文章于 2023-02-25 23:03:36 发布

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原文链接：http://www.cnblogs.com/averillzheng/p/3790281.html

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本文介绍了一种在原地修改矩阵的算法，当矩阵中的元素为0时，将该元素所在的整行和整列都置为0。算法巧妙利用了第一行和第一列作为标记空间，避免了额外的空间开销，实现了常数空间解决方案。

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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 class Solution {
	/**The algorithm is simple. but we need to pay attention to the first row and first column.
     * @author Averill Zheng
     * @version 2014-06-15
     * @since JDK 1.7
     */ 
	public void setZeroes(int[][] matrix) {
		int row = matrix.length;
		if(row > 0){
			int column = matrix[0].length;
			//find if we need to set first row or first column to be zero
			boolean setRow = false;
			boolean setColumn = false;
			for(int i = 0; i < column; ++i){
				if(matrix[0][i] == 0){
					setRow = true; 
					break;
				}
			}
			for(int i = 0; i < row; ++i){
				if(matrix[i][0] == 0){
					setColumn = true;
					break;
				}
			}
			
			for(int i = 1; i < row; ++i){
				for(int j = 1; j < column; ++j){
					if(matrix[i][j] == 0){
						matrix[i][0] = 0;
						matrix[0][j] = 0;
					}
				}
			}
			
			//set zeros;
			for(int j = 1; j < column; ++j){
				if(matrix[0][j] == 0){
					for(int i = 1; i < row; ++i)
						matrix[i][j] = 0;
				}
			}
			
			for(int i = 1; i < row; ++i){
				if(matrix[i][0] == 0){
					for(int j = 1; j < column; ++j)
						matrix[i][j] = 0;
				}
			}
			if(setRow){
				for(int i = 0; i < column; ++i)
					matrix[0][i] = 0;
			}
			
			if(setColumn){
				for(int i = 0; i < row; ++i)
					matrix[i][0] = 0;
			}
		}
    }
}

转载于:https://www.cnblogs.com/averillzheng/p/3790281.html