数组
哈希表
矩阵
题目描述
给定一个 m x n
的矩阵,如果一个元素为 0 ,则将其所在行和列的所有元素都设为 0 。请使用 原地 算法。
示例 1:
输入: matrix = [[1,1,1],[1,0,1],[1,1,1]]
输出: [[1,0,1],[0,0,0],[1,0,1]]
示例 2:
输入: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
输出: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]
提示:
m == matrix.length
n == matrix[0].length
1 <= m, n <= 200
-231 <= matrix[i][j] <= 231 - 1
进阶:
一个直观的解决方案是使用 O(m n )
的额外空间,但这并不是一个好的解决方案。
一个简单的改进方案是使用 O(m + n )
的额外空间,但这仍然不是最好的解决方案。
你能想出一个仅使用常量空间的解决方案吗?
解法
方法一:数组标记
我们分别用数组 rows
和 cols
标记待清零的行和列。
然后再遍历一遍矩阵,将 rows
和 cols
中标记的行和列对应的元素清零。
时间复杂度 $O(m\times n)$,空间复杂度 $O(m+n)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。
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13 class Solution :
def setZeroes ( self , matrix : List [ List [ int ]]) -> None :
m , n = len ( matrix ), len ( matrix [ 0 ])
rows = [ 0 ] * m
cols = [ 0 ] * n
for i , row in enumerate ( matrix ):
for j , v in enumerate ( row ):
if v == 0 :
rows [ i ] = cols [ j ] = 1
for i in range ( m ):
for j in range ( n ):
if rows [ i ] or cols [ j ]:
matrix [ i ][ j ] = 0
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22 class Solution {
public void setZeroes ( int [][] matrix ) {
int m = matrix . length , n = matrix [ 0 ] . length ;
boolean [] rows = new boolean [ m ] ;
boolean [] cols = new boolean [ n ] ;
for ( int i = 0 ; i < m ; ++ i ) {
for ( int j = 0 ; j < n ; ++ j ) {
if ( matrix [ i ][ j ] == 0 ) {
rows [ i ] = true ;
cols [ j ] = true ;
}
}
}
for ( int i = 0 ; i < m ; ++ i ) {
for ( int j = 0 ; j < n ; ++ j ) {
if ( rows [ i ] || cols [ j ] ) {
matrix [ i ][ j ] = 0 ;
}
}
}
}
}
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23 class Solution {
public :
void setZeroes ( vector < vector < int >>& matrix ) {
int m = matrix . size (), n = matrix [ 0 ]. size ();
vector < bool > rows ( m );
vector < bool > cols ( n );
for ( int i = 0 ; i < m ; ++ i ) {
for ( int j = 0 ; j < n ; ++ j ) {
if ( ! matrix [ i ][ j ]) {
rows [ i ] = 1 ;
cols [ j ] = 1 ;
}
}
}
for ( int i = 0 ; i < m ; ++ i ) {
for ( int j = 0 ; j < n ; ++ j ) {
if ( rows [ i ] || cols [ j ]) {
matrix [ i ][ j ] = 0 ;
}
}
}
}
};
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20 func setZeroes ( matrix [][] int ) {
m , n := len ( matrix ), len ( matrix [ 0 ])
rows := make ([] bool , m )
cols := make ([] bool , n )
for i , row := range matrix {
for j , v := range row {
if v == 0 {
rows [ i ] = true
cols [ j ] = true
}
}
}
for i := 0 ; i < m ; i ++ {
for j := 0 ; j < n ; j ++ {
if rows [ i ] || cols [ j ] {
matrix [ i ][ j ] = 0
}
}
}
}
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24 /**
Do not return anything, modify matrix in-place instead.
*/
function setZeroes ( matrix : number [][]) : void {
const m = matrix . length ;
const n = matrix [ 0 ]. length ;
const rows : boolean [] = new Array ( m ). fill ( false );
const cols : boolean [] = new Array ( n ). fill ( false );
for ( let i = 0 ; i < m ; ++ i ) {
for ( let j = 0 ; j < n ; ++ j ) {
if ( matrix [ i ][ j ] === 0 ) {
rows [ i ] = true ;
cols [ j ] = true ;
}
}
}
for ( let i = 0 ; i < m ; ++ i ) {
for ( let j = 0 ; j < n ; ++ j ) {
if ( rows [ i ] || cols [ j ]) {
matrix [ i ][ j ] = 0 ;
}
}
}
}
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25 /**
* @param {number[][]} matrix
* @return {void} Do not return anything, modify matrix in-place instead.
*/
var setZeroes = function ( matrix ) {
const m = matrix . length ;
const n = matrix [ 0 ]. length ;
const rows = new Array ( m ). fill ( false );
const cols = new Array ( n ). fill ( false );
for ( let i = 0 ; i < m ; ++ i ) {
for ( let j = 0 ; j < n ; ++ j ) {
if ( matrix [ i ][ j ] == 0 ) {
rows [ i ] = true ;
cols [ j ] = true ;
}
}
}
for ( let i = 0 ; i < m ; ++ i ) {
for ( let j = 0 ; j < n ; ++ j ) {
if ( rows [ i ] || cols [ j ]) {
matrix [ i ][ j ] = 0 ;
}
}
}
};
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21 public class Solution {
public void SetZeroes ( int [][] matrix ) {
int m = matrix . Length , n = matrix [ 0 ]. Length ;
bool [] rows = new bool [ m ], cols = new bool [ n ];
for ( int i = 0 ; i < m ; ++ i ) {
for ( int j = 0 ; j < n ; ++ j ) {
if ( matrix [ i ][ j ] == 0 ) {
rows [ i ] = true ;
cols [ j ] = true ;
}
}
}
for ( int i = 0 ; i < m ; ++ i ) {
for ( int j = 0 ; j < n ; ++ j ) {
if ( rows [ i ] || cols [ j ]) {
matrix [ i ][ j ] = 0 ;
}
}
}
}
}
方法二:原地标记
方法一中使用了额外的数组标记待清零的行和列,实际上我们也可以直接用矩阵的第一行和第一列来标记,不需要开辟额外的数组空间。
由于第一行、第一列用来做标记,它们的值可能会因为标记而发生改变,因此,我们需要额外的变量 $i0$, $j0$ 来标记第一行、第一列是否需要被清零。
时间复杂度 $O(m\times n)$,空间复杂度 $O(1)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。
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19 class Solution :
def setZeroes ( self , matrix : List [ List [ int ]]) -> None :
m , n = len ( matrix ), len ( matrix [ 0 ])
i0 = any ( v == 0 for v in matrix [ 0 ])
j0 = any ( matrix [ i ][ 0 ] == 0 for i in range ( m ))
for i in range ( 1 , m ):
for j in range ( 1 , n ):
if matrix [ i ][ j ] == 0 :
matrix [ i ][ 0 ] = matrix [ 0 ][ j ] = 0
for i in range ( 1 , m ):
for j in range ( 1 , n ):
if matrix [ i ][ 0 ] == 0 or matrix [ 0 ][ j ] == 0 :
matrix [ i ][ j ] = 0
if i0 :
for j in range ( n ):
matrix [ 0 ][ j ] = 0
if j0 :
for i in range ( m ):
matrix [ i ][ 0 ] = 0
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43 class Solution {
public void setZeroes ( int [][] matrix ) {
int m = matrix . length , n = matrix [ 0 ] . length ;
boolean i0 = false , j0 = false ;
for ( int j = 0 ; j < n ; ++ j ) {
if ( matrix [ 0 ][ j ] == 0 ) {
i0 = true ;
break ;
}
}
for ( int i = 0 ; i < m ; ++ i ) {
if ( matrix [ i ][ 0 ] == 0 ) {
j0 = true ;
break ;
}
}
for ( int i = 1 ; i < m ; ++ i ) {
for ( int j = 1 ; j < n ; ++ j ) {
if ( matrix [ i ][ j ] == 0 ) {
matrix [ i ][ 0 ] = 0 ;
matrix [ 0 ][ j ] = 0 ;
}
}
}
for ( int i = 1 ; i < m ; ++ i ) {
for ( int j = 1 ; j < n ; ++ j ) {
if ( matrix [ i ][ 0 ] == 0 || matrix [ 0 ][ j ] == 0 ) {
matrix [ i ][ j ] = 0 ;
}
}
}
if ( i0 ) {
for ( int j = 0 ; j < n ; ++ j ) {
matrix [ 0 ][ j ] = 0 ;
}
}
if ( j0 ) {
for ( int i = 0 ; i < m ; ++ i ) {
matrix [ i ][ 0 ] = 0 ;
}
}
}
}
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44 class Solution {
public :
void setZeroes ( vector < vector < int >>& matrix ) {
int m = matrix . size (), n = matrix [ 0 ]. size ();
bool i0 = false , j0 = false ;
for ( int j = 0 ; j < n ; ++ j ) {
if ( matrix [ 0 ][ j ] == 0 ) {
i0 = true ;
break ;
}
}
for ( int i = 0 ; i < m ; ++ i ) {
if ( matrix [ i ][ 0 ] == 0 ) {
j0 = true ;
break ;
}
}
for ( int i = 1 ; i < m ; ++ i ) {
for ( int j = 1 ; j < n ; ++ j ) {
if ( matrix [ i ][ j ] == 0 ) {
matrix [ i ][ 0 ] = 0 ;
matrix [ 0 ][ j ] = 0 ;
}
}
}
for ( int i = 1 ; i < m ; ++ i ) {
for ( int j = 1 ; j < n ; ++ j ) {
if ( matrix [ i ][ 0 ] == 0 || matrix [ 0 ][ j ] == 0 ) {
matrix [ i ][ j ] = 0 ;
}
}
}
if ( i0 ) {
for ( int j = 0 ; j < n ; ++ j ) {
matrix [ 0 ][ j ] = 0 ;
}
}
if ( j0 ) {
for ( int i = 0 ; i < m ; ++ i ) {
matrix [ i ][ 0 ] = 0 ;
}
}
}
};
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40 func setZeroes ( matrix [][] int ) {
m , n := len ( matrix ), len ( matrix [ 0 ])
i0 , j0 := false , false
for j := 0 ; j < n ; j ++ {
if matrix [ 0 ][ j ] == 0 {
i0 = true
break
}
}
for i := 0 ; i < m ; i ++ {
if matrix [ i ][ 0 ] == 0 {
j0 = true
break
}
}
for i := 1 ; i < m ; i ++ {
for j := 1 ; j < n ; j ++ {
if matrix [ i ][ j ] == 0 {
matrix [ i ][ 0 ], matrix [ 0 ][ j ] = 0 , 0
}
}
}
for i := 1 ; i < m ; i ++ {
for j := 1 ; j < n ; j ++ {
if matrix [ i ][ 0 ] == 0 || matrix [ 0 ][ j ] == 0 {
matrix [ i ][ j ] = 0
}
}
}
if i0 {
for j := 0 ; j < n ; j ++ {
matrix [ 0 ][ j ] = 0
}
}
if j0 {
for i := 0 ; i < m ; i ++ {
matrix [ i ][ 0 ] = 0
}
}
}
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32 /**
Do not return anything, modify matrix in-place instead.
*/
function setZeroes ( matrix : number [][]) : void {
const m = matrix . length ;
const n = matrix [ 0 ]. length ;
const i0 = matrix [ 0 ]. includes ( 0 );
const j0 = matrix . map ( row => row [ 0 ]). includes ( 0 );
for ( let i = 1 ; i < m ; ++ i ) {
for ( let j = 1 ; j < n ; ++ j ) {
if ( matrix [ i ][ j ] === 0 ) {
matrix [ i ][ 0 ] = 0 ;
matrix [ 0 ][ j ] = 0 ;
}
}
}
for ( let i = 1 ; i < m ; ++ i ) {
for ( let j = 1 ; j < n ; ++ j ) {
if ( matrix [ i ][ 0 ] === 0 || matrix [ 0 ][ j ] === 0 ) {
matrix [ i ][ j ] = 0 ;
}
}
}
if ( i0 ) {
matrix [ 0 ]. fill ( 0 );
}
if ( j0 ) {
for ( let i = 0 ; i < m ; ++ i ) {
matrix [ i ][ 0 ] = 0 ;
}
}
}
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41 /**
* @param {number[][]} matrix
* @return {void} Do not return anything, modify matrix in-place instead.
*/
var setZeroes = function ( matrix ) {
const m = matrix . length ;
const n = matrix [ 0 ]. length ;
let i0 = matrix [ 0 ]. some ( v => v == 0 );
let j0 = false ;
for ( let i = 0 ; i < m ; ++ i ) {
if ( matrix [ i ][ 0 ] == 0 ) {
j0 = true ;
break ;
}
}
for ( let i = 1 ; i < m ; ++ i ) {
for ( let j = 1 ; j < n ; ++ j ) {
if ( matrix [ i ][ j ] == 0 ) {
matrix [ i ][ 0 ] = 0 ;
matrix [ 0 ][ j ] = 0 ;
}
}
}
for ( let i = 1 ; i < m ; ++ i ) {
for ( let j = 1 ; j < n ; ++ j ) {
if ( matrix [ i ][ 0 ] == 0 || matrix [ 0 ][ j ] == 0 ) {
matrix [ i ][ j ] = 0 ;
}
}
}
if ( i0 ) {
for ( let j = 0 ; j < n ; ++ j ) {
matrix [ 0 ][ j ] = 0 ;
}
}
if ( j0 ) {
for ( let i = 0 ; i < m ; ++ i ) {
matrix [ i ][ 0 ] = 0 ;
}
}
};
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37 public class Solution {
public void SetZeroes ( int [][] matrix ) {
int m = matrix . Length , n = matrix [ 0 ]. Length ;
bool i0 = matrix [ 0 ]. Contains ( 0 ), j0 = false ;
for ( int i = 0 ; i < m ; ++ i ) {
if ( matrix [ i ][ 0 ] == 0 ) {
j0 = true ;
break ;
}
}
for ( int i = 1 ; i < m ; ++ i ) {
for ( int j = 1 ; j < n ; ++ j ) {
if ( matrix [ i ][ j ] == 0 ) {
matrix [ i ][ 0 ] = 0 ;
matrix [ 0 ][ j ] = 0 ;
}
}
}
for ( int i = 1 ; i < m ; ++ i ) {
for ( int j = 1 ; j < n ; ++ j ) {
if ( matrix [ i ][ 0 ] == 0 || matrix [ 0 ][ j ] == 0 ) {
matrix [ i ][ j ] = 0 ;
}
}
}
if ( i0 ) {
for ( int j = 0 ; j < n ; ++ j ) {
matrix [ 0 ][ j ] = 0 ;
}
}
if ( j0 ) {
for ( int i = 0 ; i < m ; ++ i ) {
matrix [ i ][ 0 ] = 0 ;
}
}
}
}
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