[LeetCode] 73. Set Matrix Zeroes 矩阵赋零
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.
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?
给一个m x n的矩阵，如果一个元素是0，就把它所在的行和列都设成0，用in place做。
解法1: 新建一个矩阵，然后一行一行的扫，只要有0，就将新建的矩阵的对应行全赋0，行扫完再扫列，然后把更新完的矩阵赋给matrix。空间复杂度为O(mn)。
解法2: 用一个长度为m的一维数组记录各行中是否有0，用一个长度为n的一维数组记录各列中是否有0，最后直接更新matrix数组即可。空间复杂度为O(m+n)，
解法3: 这道题要求用常数级空间复杂度O(1)，不能新建数组，就用原数组的第一行第一列来记录各行各列是否有0.
- 先扫描第一行第一列，如果有0，则将各自的flag设置为true
- 然后扫描除去第一行第一列的整个数组，如果有0，则将对应的第一行和第一列的数字赋0
- 再次遍历除去第一行第一列的整个数组，如果对应的第一行和第一列的数字有一个为0，则将当前值赋0
- 最后根据第一行第一列的flag来更新第一行第一列
时间复杂度是O(m*n), 三种方法都一样，需要进行两次扫描，一次确定行列置0情况，一次对矩阵进行实际的置0操作。
Java:
void setZeroes(vector<vector<int> > &matrix) {
int col0 = 1, rows = matrix.size(), cols = matrix[0].size();
for (int i = 0; i < rows; i++) {
if (matrix[i][0] == 0) col0 = 0;
for (int j = 1; j < cols; j++)
if (matrix[i][j] == 0)
matrix[i][0] = matrix[0][j] = 0;
}
for (int i = rows - 1; i >= 0; i--) {
for (int j = cols - 1; j >= 1; j--)
if (matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
if (col0 == 0) matrix[i][0] = 0;
}
}
Java:
public void setZeroes(int[][] matrix) {
boolean fr = false,fc = false;
for(int i = 0; i < matrix.length; i++) {
for(int j = 0; j < matrix[0].length; j++) {
if(matrix[i][j] == 0) {
if(i == 0) fr = true;
if(j == 0) fc = true;
matrix[0][j] = 0;
matrix[i][0] = 0;
}
}
}
for(int i = 1; i < matrix.length; i++) {
for(int j = 1; j < matrix[0].length; j++) {
if(matrix[i][0] == 0 || matrix[0][j] == 0) {
matrix[i][j] = 0;
}
}
}
if(fr) {
for(int j = 0; j < matrix[0].length; j++) {
matrix[0][j] = 0;
}
}
if(fc) {
for(int i = 0; i < matrix.length; i++) {
matrix[i][0] = 0;
}
}
}
Java:
public class Solution {
public void setZeroes(int[][] matrix) {
boolean firstRowZero = false;
boolean firstColumnZero = false;
//set first row and column zero or not
for(int i=0; i<matrix.length; i++){
if(matrix[i][0] == 0){
firstColumnZero = true;
break;
}
}
for(int i=0; i<matrix[0].length; i++){
if(matrix[0][i] == 0){
firstRowZero = true;
break;
}
}
//mark zeros on first row and column
for(int i=1; i<matrix.length; i++){
for(int j=1; j<matrix[0].length; j++){
if(matrix[i][j] == 0){
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
//use mark to set elements
for(int i=1; i<matrix.length; i++){
for(int j=1; j<matrix[0].length; j++){
if(matrix[i][0] == 0 || matrix[0][j] == 0){
matrix[i][j] = 0;
}
}
}
//set first column and row
if(firstColumnZero){
for(int i=0; i<matrix.length; i++)
matrix[i][0] = 0;
}
if(firstRowZero){
for(int i=0; i<matrix[0].length; i++)
matrix[0][i] = 0;
}
}
}
Java:
public void setZeroes(int[][] matrix) {
if(matrix==null || matrix.length==0 || matrix[0].length==0)
return;
boolean rowFlag = false;
boolean colFlag = false;
for(int i=0;i<matrix.length;i++)
{
if(matrix[i][0]==0)
{
colFlag = true;
break;
}
}
for(int i=0;i<matrix[0].length;i++)
{
if(matrix[0][i]==0)
{
rowFlag = true;
break;
}
}
for(int i=1;i<matrix.length;i++)
{
for(int j=1;j<matrix[0].length;j++)
{
if(matrix[i][j]==0)
{
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
for(int i=1;i<matrix.length;i++)
{
for(int j=1;j<matrix[0].length;j++)
{
if(matrix[i][0]==0 || matrix[0][j]==0)
matrix[i][j] = 0;
}
}
if(colFlag)
{
for(int i=0;i<matrix.length;i++)
{
matrix[i][0] = 0;
}
}
if(rowFlag)
{
for(int i=0;i<matrix[0].length;i++)
{
matrix[0][i] = 0;
}
}
}
Python:
class Solution:
# @param matrix, a list of lists of integers
# RETURN NOTHING, MODIFY matrix IN PLACE.
def setZeroes(self, matrix):
first_col = reduce(lambda acc, i: acc or matrix[i][0] == 0, xrange(len(matrix)), False)
first_row = reduce(lambda acc, j: acc or matrix[0][j] == 0, xrange(len(matrix[0])), False)
for i in xrange(1, len(matrix)):
for j in xrange(1, len(matrix[0])):
if matrix[i][j] == 0:
matrix[i][0], matrix[0][j] = 0, 0
for i in xrange(1, len(matrix)):
for j in xrange(1, len(matrix[0])):
if matrix[i][0] == 0 or matrix[0][j] == 0:
matrix[i][j] = 0
if first_col:
for i in xrange(len(matrix)):
matrix[i][0] = 0
if first_row:
for j in xrange(len(matrix[0])):
matrix[0][j] = 0
Python:
class Solution:
# @param {integer[][]} matrix
# @return {void} Do not return anything, modify matrix in-place instead.
def setZeroes(self, matrix):
m = len(matrix)
if m == 0:
return
n = len(matrix[0])
row_zero = False
for i in range(m):
if matrix[i][0] == 0:
row_zero = True
col_zero = False
for j in range(n):
if matrix[0][j] == 0:
col_zero = True
for i in range(1, m):
for j in range(1, n):
if matrix[i][j] == 0:
matrix[i][0] = 0
matrix[0][j] = 0
for i in range(1, m):
if matrix[i][0] == 0:
for j in range(1, n):
matrix[i][j] = 0
for j in range(1, n):
if matrix[0][j] == 0:
for i in range(1, m):
matrix[i][j] = 0
if col_zero:
for j in range(n):
matrix[0][j] = 0
if row_zero:
for i in range(m):
matrix[i][0] = 0
C++:
class Solution {
public:
void setZeroes(vector<vector<int> > &matrix) {
if (matrix.empty() || matrix[0].empty()) return;
int m = matrix.size(), n = matrix[0].size();
bool rowZero = false, colZero = false;
for (int i = 0; i < m; ++i) {
if (matrix[i][0] == 0) colZero = true;
}
for (int i = 0; i < n; ++i) {
if (matrix[0][i] == 0) rowZero = true;
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
if (matrix[i][j] == 0) {
matrix[0][j] = 0;
matrix[i][0] = 0;
}
}
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
if (matrix[0][j] == 0 || matrix[i][0] == 0) {
matrix[i][j] = 0;
}
}
}
if (rowZero) {
for (int i = 0; i < n; ++i) matrix[0][i] = 0;
}
if (colZero) {
for (int i = 0; i < m; ++i) matrix[i][0] = 0;
}
}
};
