leetcode 73. Set Matrix Zeroes

文章目录

• C++
• Java
• Python
• Go

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

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?

Example 1:

Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
Output: [[1,0,1],[0,0,0],[1,0,1]]
Example 2:

Input: matrix = [[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]]

Constraints:

m == matrix.length
n == matrix[0].length
1 <= m, n <= 200
-10^9 <= matrix[i][j] <= 10^9

C++

• $O(MN)$ 时间复杂度，$O(1)$ 空间复杂度
• 思路是将第一行和第一列作为标志
• 打完标记后再逆着根据标记置零
• 细节还是有的，比如要选择把第一行或者第一列空出来，然后选择对应的第一列的标记或者第一行的标记
• 代码中的写法是空出了第一列，然后用一个额外变量来记录第一列是否置零

class Solution {
public:
    void setZeroes(vector<vector<int>>& a) {
        int n=a.size(),m=a[0].size();
        bool col=false;
        for(int i=0;i<n;i++){
            if(!a[i][0])col=true;
            for(int j=1;j<m;j++)
                if(!a[i][j])
                    a[i][0]=a[0][j]=0;
        }
        for(int i=n-1;i>=0;i--){
            for(int j=m-1;j>=1;j--)
                if(!a[i][0]||!a[0][j])
                    a[i][j]=0;
            if(col)a[i][0]=0;
        }
    }
};

Java

class Solution {
    public void setZeroes(int[][] a) {
        int n=a.length,m=a[0].length;
        boolean col=false;
        for(int i=0;i<n;i++){
            if(a[i][0]==0)col=true;
            for(int j=1;j<m;j++)
                if(a[i][j]==0)
                    a[i][0]=a[0][j]=0;
        }
        for(int i=n-1;i>=0;i--){
            for(int j=m-1;j>=1;j--)
                if(a[i][0]==0||a[0][j]==0)
                    a[i][j]=0;
            if(col)a[i][0]=0;
        }
    }
}

Python

class Solution:
    def setZeroes(self, a: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        n,m,col=len(a),len(a[0]),False
        for i in range(n):
            if a[i][0]==0:
                col=True
            for j in range(1,m):
                if a[i][j]==0:
                    a[i][0]=a[0][j]=0
        for i in range(n-1,-1,-1):
            for j in range(m-1,0,-1):
                if a[i][0]==0 or a[0][j]==0:
                    a[i][j]=0
            if col:
                a[i][0]=0
    

Go

func setZeroes(a [][]int)  {
    n,m,col:=len(a),len(a[0]),false
    for i:=0;i<n;i++ {
        if a[i][0]==0 {
            col=true
        }
        for j:=1;j<m;j++ {
            if a[i][j]==0 {
                a[i][0],a[0][j]=0,0
            }
        }
    }
    for i:=n-1;i>=0;i-- {
        for j:=m-1;j>=1;j-- {
            if a[i][0]==0 || a[0][j]==0 {
                a[i][j]=0
            }
        }
        if col {
            a[i][0]=0
        }
    }
}