[Leetcode] set matrix zeroes solution report

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[Question]

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

Click to show follow up.

Follow up:

Did you use extra space?
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.
Cocould you devise a constant space solution?

[Idea]

After reading the following requirements, I first thought of the space of O (m + n), that is, adding a row and a column to mark which row and column have 0; instead, I don't know how to use the space of O (Mn.

Next I thought about the space of O (1). I thought about it a little and thought of the answer, that is, compared to the space of O (m + n), no additional row or column is added, the first row and the first column are used to store which row and which column has 0. Of course, it is troublesome to save the original information in the first column of the first row.

During the first two writes, because the first column of the First row is not properly processed, WA is caused.

After reading the answer from the Internet, I am very happy to see other people.

[Java code, O (1) Space]

Public class solution {public void setzeroes (INT [] [] matrix) {int M = matrix. length; int n = matrix [0]. length; int I, j; // indicates whether the first row and the first column have 0 Boolean firstrow = false, firstcol = false; For (j = 0; j <N; j ++) {If (0 = matrix [0] [J]) {firstrow = true; break ;}} for (I = 0; I <m; I ++) {If (0 = matrix [I] [0]) {firstcol = true; break ;}// traverse the second column of the Second row, if 0 is encountered, set the first value of the row and column to 0 for (I = 1; I <m; I ++) {for (j = 1; j <n; j ++) {If (0 = matrix [I] [J]) {matrix [I] [0] = 0; matrix [0] [J] = 0 ;}}// set the row 0 in the first column to 0, set the column where the first row 0 is located to 0 for (I = 1; I <m; I ++) {If (0 = matrix [I] [0]) {for (j = 1; j <n; j ++) {matrix [I] [J] = 0 ;}}for (j = 1; j <N; j ++) {If (0 = matrix [0] [J]) {for (I = 1; I <m; I ++) {matrix [I] [J] = 0 ;}}// determines whether the first row and the first column are set to 0 if (firstrow) {for (j = 0; j <n; j ++) {matrix [0] [J] = 0 ;}} if (firstcol) {for (I = 0; I <m; I ++) {matrix [I] [0] = 0 ;}}}}

[Leetcode] set matrix zeroes solution report