[LeetCode] 73. Set Matrix Zeroes 解题思路

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?

问题：给定一个矩阵，当矩阵一个元素为 0 ，将该元素的当前行，当前列都设为 0 。

我的方案满足补充内容的第二点，使用 O(m + n) 额外空间。

• 将矩阵中所有 0 元素的所在行，列记录下来。
• 将记录下来的行号，整行设为0，将记录下来的列号，整列设为0.

 1 void setZeroes(vector<vector<int>>& matrix) {
 2
 3     unordered_set<int> seti;
 4     unordered_set<int> setk;
 5
 6     for (int i = 0 ; i < matrix.size(); i++) {
 7         for (int k = 0 ; k < matrix[0].size(); k++) {
 8             if (matrix[i][k] == 0) {
 9                 seti.insert(i);
10                 setk.insert(k);
11             }
12         }
13     }
14
15     unordered_set<int>::iterator s_iter;
16     for (s_iter = seti.begin(); s_iter != seti.end(); s_iter++) {
17         for (int k = 0 ; k < matrix[0].size(); k++) {
18             matrix[*s_iter][k] = 0;
19         }
20     }
21
22     for (s_iter = setk.begin(); s_iter != setk.end(); s_iter++) {
23         for (int i = 0; i < matrix.size(); i++) {
24             matrix[i][*s_iter] = 0;
25         }
26     }
27 }