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?
public class Solution {
public void setZeroes(int[][] matrix) {
int m=matrix.length;
if(m==0)
return;
int n=matrix[0].length;
if(n==0)
return;
boolean preZeros=false;
boolean curZeros=false;
for(int i=0;i<m;i++)
{
//test if this row has zeros.
for(int j=0;j<n;j++)
{
if(matrix[i][j]==0)
{
for(int k=0;k<i;k++)
matrix[k][j]=0;
curZeros=true;
}
}
//fill the zeros along coloum.
if(i>0)
{
for(int j=0;j<n;j++)
{
if(matrix[i-1][j]==0)
{
matrix[i][j]=0;
}
}
}
//if pre row has zeros, fill zeros with that row.
if(preZeros)
{
for(int j=0;j<n;j++)
{
matrix[i-1][j]=0;
}
}
preZeros=curZeros;
curZeros=false;
}
if(preZeros)
{
for(int j=0;j<n;j++)
{
matrix[m-1][j]=0;
}
}
}
}
将矩阵中值为0的元素所在的行和列设置为0, in-place O(1)space O(mn) time
原文:http://blog.csdn.net/smartxxyx/article/details/44964611