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?
题意:给定m*n的矩阵,如果某个元素是0,使其同行,同列的所有元素为0
要求使用O(m+n)空间复制度,甚至常数空间
解法1:首先遍历第一行,如果找到0,做标记,说明第一行要清0
遍历第一列,如果找到0,做标记,说明第一列要清0
然后从(1,1)开始遍历矩阵,找到0,将对应的第一行,第一列的元素设置为0,用于标记该行列应该清0
最后,根据第一行,第一列,0的情况,清0
public class Solution { public void setZeroes( int [][] matrix) { int rows = matrix.length; int cols = matrix[ 0 ].length; boolean firstRow= false ,firstCol= false ; for ( int i= 0 ;i<rows;i++){ if (matrix[i][ 0 ]== 0 ){ firstRow = true ; break ; } } for ( int i= 0 ;i<cols;i++){ if (matrix[ 0 ][i]== 0 ){ firstCol = true ; break ; } } for ( int i= 1 ;i<rows;i++){ for ( int j= 1 ;j<cols;j++){ if (matrix[i][j]== 0 ){ matrix[i][0 ] = 0 ; matrix[0 ][j] = 0 ; } } } for ( int i= 1 ;i<rows;i++){ if (matrix[i][ 0 ]== 0 ){ for ( int j= 1 ;j<cols;j++){ matrix[i][j] = 0 ; } } } for ( int j= 1 ;j<cols;j++){ if (matrix[ 0 ][j]== 0 ){ for ( int i= 1 ;i<rows;i++){ matrix[i][j] = 0 ; } } } if (firstRow) for ( int i= 0 ;i<rows;i++) matrix[i][ 0 ] = 0 ; if (firstCol) for ( int j= 0 ;j<cols;j++) matrix[ 0 ][j] = 0 ; } }
原文链接http://blog.youkuaiyun.com/crazy__chen/article/details/46413689