Leetcode [73] (Java): Set Matrix Zeroes (matrix 0)

title : Matrix 0

Difficulty : Easy

topic content :

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

translation :

Given an m x n matrix, if an element is 0, its entire row and column is set to 0.

Requirement: Place 0.

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]]

my idea : Because the requirement is in place, so cannot create a new matrix, and then meet 0 will be placed in the row all 0

So it's all over again, and all 0 of the rows are labeled with list<integer[]>, the number No. 0 is row, and the 1th is the column.

Then loop this list, call the 0 method respectively.

My Code :

1      Public voidSetzeroes (int[] matrix) {2         if(Matrix.length = = 0 | | matrix[0].length = = 0)3             return;4list<integer[]> list =NewArraylist<integer[]>();5          for(inti = 0; i < matrix.length; i++) {6              for(intj = 0; J < Matrix[0].length; J + +) {7                 if(Matrix[i][j] = = 0) {8List.add (Newinteger[]{i,j});9                 }Ten             } One         } A          for(integer[] x:list) { -Setzero (Matrix, x[0], x[1]); -         } the     } -      -      Public voidSetzero (int[] Matrix,intIintj) { -          for(intn = 0; n < matrix.length; n++) { +MATRIX[N][J] = 0; -         } +          for(intm = 0; M < matrix[0].length; m++) { AMatrix[i][m] = 0; at         } -}

my complexity : O (m*n) + O ((m+n) *x) x is the number of 0 in the matrix

problems in the programming process :

1. No

(integer[in code] can be replaced with int[]

Answer code :

1      Public voidSetzeroes (int[] matrix) {2         if(Matrix.length = = 0 | | matrix[0].length = = 0)3             return;4         Booleanrow =false, col =false;5          for(inti = 0; i < matrix.length; i++) {6              for(intj = 0; J < Matrix[0].length; J + +) {7                 if(Matrix[i][j] = = 0) {8                     if(i = = 0) row =true;9                     if(j = = 0) col =true;TenMatrix[i][0] = 0; OneMATRIX[0][J] = 0; A                 } -             } -         } the          for(inti = 1; i < matrix.length; i++) {//Note that this should be starting from 1, because the "0,0" if the 0 is also judged by the No. 0 column (the flag column) all set to 0, then the entire matrix will be 0 -             if(matrix[i][0] = = 0) { -                  for(intj = 0; J < Matrix[0].length; J + +) { -MATRIX[I][J] = 0; +                 } -             } +         } A          for(intj = 1; J < Matrix[0].length; J + +) {//  at             if(Matrix[0][j] = = 0) { -                  for(inti = 0; i < matrix.length; i++) { -MATRIX[I][J] = 0; -                 } -             } -         } in         if(row) { -              for(intj = 0; J < Matrix[0].length; J + +) { toMATRIX[0][J] = 0; +             } -         } the         if(col) { *              for(inti = 0; i < matrix.length; i++) { $Matrix[i][0] = 0;Panax Notoginseng             } -         } the}

complexity of the answer : O (m*n)

The answer : using rows No. 0 and No. 0 as the "flag bit" and setting two Boolean variables to record whether the "flag bit" needs to be set 0, this method compared to my method in the last 0 to avoid the repetition of 0 of the action, the original matrix of more than 0, Only O (m*n) secondary 0 is required

Leetcode [73] (Java): Set Matrix Zeroes (matrix 0)