73. Set Matrix Zeroes
Given an m x n
integer matrix matrix
, if an element is 0
, set its entire row and column to 0
's, and return the matrix.
You must do it in place.
Example 1:
Input: matrix = [[1,1,1],[1,0,1],[1,1,1]] Output: [[1,0,1],[0,0,0],[1,0,1]]
Example 2:
Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]] Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
Constraints:
m == matrix.length
n == matrix[0].length
1 <= m, n <= 200
-231 <= matrix[i][j] <= 231 - 1
Follow up:
- A straightforward 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?
The solution to the above question
class Solution {
public:
void setZeroes(vector<vector<int>>& v) {
bool row=false;
bool col=false;
for(int i=0; i<v.size(); i++){
for(int j=0; j<v[0].size(); j++){
if(v[i][j]==0){
v[0][j]=0;
v[i][0]=0;
if(i==0){
row=true;
}
if(j==0){
col=true;
}
}
}
}
for(int i=1; i<v[0].size(); i++){
if(v[0][i]==0){
for(int j=0; j<v.size(); j++){
v[j][i]=0;
}
}
}
for(int i=1; i<v.size(); i++){
if(v[i][0]==0){
for(int j=0; j<v[0].size(); j++){
v[i][j]=0;
}
}
}
if(row){
for(int i=0; i<v[0].size(); i++){
v[0][i]=0;
}
}
if(col){
for(int i=0; i<v.size(); i++){
v[i][0]=0;
}
}
for(int i=0; i<v.size(); i++){
for(int j=0; j<v[0].size(); j++){
{
cout<<v[i][j]<<" ";
}
}
cout<<endl;
}
}
};
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