Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
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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?
找出矩阵中的0点，并将所在行列都置0
具体要求为使用较小的空间，开始m n的做法
代码语言：javascript复制class Solution {
public:
void setZeroes(vector<vector<int>>& matrix) {
vector<int> x;
for(int i=0;i<matrix.size();i )
for(int j=0;j<matrix[0].size();j )
if(matrix[i][j]==0)
{
x.push_back(i);
x.push_back(j);
}
for(int i=0;i<x.size();i =2)
{
for(int j=0;j<matrix.size();j ) matrix[j][x[i 1]]=0;
for(int j=0;j<matrix[0].size();j ) matrix[x[i]][j]=0;
}
}
};在discuss中看到的O(1)做法，将是否有0存在每一行每一列的第一个位置，
因为行列会交叉，因而会当左上角为0时会不知道到底是行还是列，
所以引入col0记录，col0为0表示是第一列产生的0
代码语言：javascript复制void setZeroes(vector<vector<int> > &matrix) {
int col0 = 1, rows = matrix.size(), cols = matrix[0].size();
for (int i = 0; i < rows; i ) {
if (matrix[i][0] == 0) col0 = 0;
for (int j = 1; j < cols; j )
if (matrix[i][j] == 0)
matrix[i][0] = matrix[0][j] = 0;
}
for (int i = rows - 1; i >= 0; i--) {
for (int j = cols - 1; j >= 1; j--)
if (matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
if (col0 == 0) matrix[i][0] = 0;
}
}
