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?

1,O(mn),即用的一个相同大小的矩阵来记录那个位置有0.

2,O(m + n),就是加一行一列来标记哪行哪列有0。

3,常数空间,即考虑不使用额外的空间,可以把第一行与第一列作为标记行与标记列,但是得先确定第一行与第一列本身要不要设为0。

代码非常简单:

class Solution {
public:
void setZeroes(vector<vector<int>>& matrix) {
if (matrix.size() < ) return;
int row = matrix.size(), col = matrix[].size();
bool r0 = false, c0 = false;
for (int i = ; i < row; ++i) {
if (matrix[i][] == ) {
c0 = true; break;
}
}
for (int j = ; j < col; ++j) {
if (matrix[][j] == ) {
r0 = true; break;
}
}
for (int i = ; i < row; ++i) {
for (int j = ; j < col; ++j) {
matrix[i][] = (matrix[i][j] == ) ? : matrix[i][];
matrix[][j] = (matrix[i][j] == ) ? : matrix[][j];
}
}
for (int i = ; i < row; ++i) {
for (int j = ; j < col; ++j) {
matrix[i][j] = (matrix[i][] == ) ? : matrix[i][j];
matrix[i][j] = (matrix[][j] == ) ? : matrix[i][j];
}
}
for (int i = ; i < row && c0; ++i) matrix[i][] = ;
for (int j = ; j < col && r0; ++j) matrix[][j] = ;
}
};

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