website/content.en/ChapterFour/0001~0099/0073.Set-Matrix-Zeroes.md
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:
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.lengthn == matrix[0].length1 <= m, n <= 2002^31 <= matrix[i][j] <= 2^31 - 1Given an m x n matrix, if an element is 0, set all elements in its row and column to 0. Please use an in-place algorithm.
package leetcode
func setZeroes(matrix [][]int) {
if len(matrix) == 0 || len(matrix[0]) == 0 {
return
}
isFirstRowExistZero, isFirstColExistZero := false, false
for i := 0; i < len(matrix); i++ {
if matrix[i][0] == 0 {
isFirstColExistZero = true
break
}
}
for j := 0; j < len(matrix[0]); j++ {
if matrix[0][j] == 0 {
isFirstRowExistZero = true
break
}
}
for i := 1; i < len(matrix); i++ {
for j := 1; j < len(matrix[0]); j++ {
if matrix[i][j] == 0 {
matrix[i][0] = 0
matrix[0][j] = 0
}
}
}
// Set all rows [1:] to 0
for i := 1; i < len(matrix); i++ {
if matrix[i][0] == 0 {
for j := 1; j < len(matrix[0]); j++ {
matrix[i][j] = 0
}
}
}
// Set all columns [1:] to 0
for j := 1; j < len(matrix[0]); j++ {
if matrix[0][j] == 0 {
for i := 1; i < len(matrix); i++ {
matrix[i][j] = 0
}
}
}
if isFirstRowExistZero {
for j := 0; j < len(matrix[0]); j++ {
matrix[0][j] = 0
}
}
if isFirstColExistZero {
for i := 0; i < len(matrix); i++ {
matrix[i][0] = 0
}
}
}