问题描述:
Given a m x 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.
Could you devise a constant space solution?
我的空间复杂度是O(m + n),时间复杂度惨不忍睹
思路是这样的,如果某一点为0,则把该点的x放在一个vector中,把y放在另一个vector中,然后对这两个vector进行去重操作,最后把矩阵中x和y在这两个vector中的点全部赋为0
上代码
class Solution {
public:
void setZeroes(vector<vector<int> > &matrix) {
int m = matrix.size();
int n = matrix[0].size();
vector<int> x_m;
vector<int> y_n;
for(int x=0;x<m;x++)
{
for(int y=0;y<n;y++)
{
if(matrix[x][y]==0)
{
x_m.push_back(x);
y_n.push_back(y);
}
}
}
x_m.erase(unique(x_m.begin(), x_m.end()),x_m.end());
y_n.erase(unique(y_n.begin(), y_n.end()),y_n.end());
for(int x=0;x<x_m.size();x++)
{
for(int i=0;i<matrix[0].size();i++)
{
matrix[x_m[x]][i]=0;
}
}
for(int y=0;y<y_n.size();y++)
{
for(int i=0;i<matrix.size();i++)
{
matrix[i][y_n[y]]=0;
}
}
}
};