【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
Set Matrix Zeroes
Example 1:
Input:
[
[1,1,1],
[1,0,1],
[1,1,1]
]
Output:
[
[1,0,1],
[0,0,0],
[1,0,1]
]
Example 2:
Input:
[
[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]
]
Follow up:
-
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?
//将矩阵中为0的元素其整行整列均置为0，要求in-place解决
//方法：第一次扫描，用首行首列存储状态，第二次扫描，根据这些状态把相应位置置0
//O(mn),O(1)
class Solution
{
public:
void setZeroes(vector<vector<int>>& a)
{
int rows = a.size();
int cols = a[0].size();
bool fr = false; //用于表示首行是否存在0元素的标识变量
bool fc = false;
for(int i = 0; i<rows; i++)
{
for(int j = 0; j<cols; j++)
{
if(a[i][j] == 0)
{
a[i][0] = a[0][j] = 0; //当某个元素为0时，将行首和列首元素置0
if(i == 0) fr = true; //第一行有0元素
if(j == 0) fc = true; //第一列有0元素
}
}
}
for(int i = 1; i<rows; i++) //从a11开始扫描，以免破坏首行首列存储信息
{
for(int j = 1; j<cols; j++)
{
if(a[i][0] == 0 || a[0][j] == 0) a[i][j] = 0;
}
}
if(fr)
{
for(int j = 0; j<cols;j++) a[0][j] = 0; //将第一行置0
}
if(fc)
{
for(int i = 0; i<rows; i++) a[i][0] = 0; //将第一列置0
}
}
};