Given a m n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
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.
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:
这题要求的是空间复杂度。
思路1,O(m+n)的方法,设置连个bool数组,记录每行和每列是否存在0;
思路2,O(1),复用第一行第一列。
#include <iostream>
#include <vector>
using namespace std;
//此算法的空间复杂度是常数级别。
class soultion{
public:
void setZeros(vector<vector<int>> &matrix)
{
int m = matrix.size();
int n = matrix[0].size();
bool first_row_zero = false;
bool first_col_zero = false;
for(int i=0;i<n;i++)
{
if(matrix[0][i] == 0)
first_row_zero = true;
break;
}
for(int i=0;i<m;i++)
{
if(matrix[i][0] == 0)
first_col_zero = true;
break;
}
for(int i=1;i<m;i++)
for(int j=1;j<n;j++)
if(matrix[i][j] == 0){
matrix[0][j]=0;
matrix[i][0]=0;
}
//解决除了第一行和第一列之外
for(int i=1;i<m;i++)
for(int j=1;j<n;j++)
if(matrix[i][0]==0 || matrix[0][j]==0)
matrix[i][j] = 0;
//解决第一行,第一列
if(first_row_zero)
for(int i=0;i<n;i++)
matrix[0][i]=0;
if(first_col_zero)
for(int i=0;i<m;i++)
matrix[i][0]=0;
}
}