[LeetCode]Set Matrix Zeroes

最新推荐文章于 2025-09-04 16:11:29 发布

weixin_30887919

最新推荐文章于 2025-09-04 16:11:29 发布

阅读量43

点赞数

CC 4.0 BY-SA版权

文章标签：数据结构与算法开发工具

原文链接：http://www.cnblogs.com/skycore/p/4884001.html

本文详细介绍了如何通过定义两个数组来记录矩阵中某一行或某一列是否包含零元素，并据此将相应行和列置零的操作。此算法在空间复杂度上进行了优化，避免了使用额外的存储空间。

题目描述:(链接)

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.

click to show follow up.

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?

解题思路：

定义两个数组分别记录某一行或者某一列是否包含0。

 1 class Solution {
 2 public:
 3     void setZeroes(vector<vector<int>>& matrix) {
 4         int rows = matrix.size();
 5         int cols = matrix[0].size();
 6         
 7         vector<bool> rowsVec(rows, false);
 8         vector<bool> colsVec(cols, false);
 9         
10         for (int i = 0; i < rows; ++i) {
11             for (int j = 0; j < cols; ++j) {
12                 if (matrix[i][j] == 0) {
13                     rowsVec[i] = true;
14                     colsVec[j] = true;
15                     continue;
16                 }
17             }
18         }
19         
20         for (int i = 0; i < rows; ++i) {
21             if (rowsVec[i]) {
22                 fill(&matrix[i][0], &matrix[i][0] + cols, 0);
23             }
24         }
25         
26         for (int j = 0; j < cols; ++j) {
27             if (colsVec[j]) {
28                 for (int i = 0; i < rows; ++i) {
29                     matrix[i][j] = 0;
30                 }
31             }
32         }
33     }
34 };