73. Set Matrix Zeroes - Medium
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?
把第一行和第一列的元素当作flag,表示这一行/列有无0
首先扫描matrix,如果有元素为0,就把该元素所在的行、列的第一个元素设为0,作为稍后整体设零的flag
注意第一行和第一列的处理!!如果第一行/列有0,用另外的flag表示,最后再处理这两部分。如果一开始就设0的话,整个矩阵都会变成0
扫描第一行,如果有0,就把当前列全部设为0;扫描第一列,如果有0,就把当前行全部设为0
最后处理第一行/列,如果有0的话
time: O(M * N), space: O(1)
class Solution { public void setZeroes(int[][] matrix) { if(matrix == null || matrix.length == 0) return; int m = matrix.length, n = matrix[0].length; boolean rowHasZero = false, colHasZero = false; for(int i = 0; i < m; i++) { for(int j = 0; j < n; j++) { if(matrix[0][j] == 0) rowHasZero = true; if(matrix[i][0] == 0) colHasZero = true; } } for(int i = 0; i < m; i++) { for(int j = 0; j < n; j++) { if(matrix[i][j] == 0) { matrix[i][0] = 0; matrix[0][j] = 0; } } } for(int i = 1; i < m; i++) { if(matrix[i][0] == 0) { for(int j = 1; j < n; j++) matrix[i][j] = 0; } } for(int j = 1; j < n; j++) { if(matrix[0][j] == 0) { for(int i = 1; i < m; i++) matrix[i][j] = 0; } } if(colHasZero) { for(int i = 0; i < m; i++) matrix[i][0] = 0; } if(rowHasZero) { for(int j = 0; j < n; j++) matrix[0][j] = 0; } } }
改进:少一个pass
class Solution { public void setZeroes(int[][] matrix) { if(matrix == null || matrix.length == 0) return; int m = matrix.length, n = matrix[0].length; boolean rowHasZero = false, colHasZero = false; for(int i = 0; i < m; i++) { for(int j = 0; j < n; j++) { if(matrix[i][j] == 0) { matrix[i][0] = 0; matrix[0][j] = 0; if(j == 0) colHasZero = true; if(i == 0) rowHasZero = true; } } } for(int i = 1; i < m; i++) { if(matrix[i][0] == 0) { for(int j = 1; j < n; j++) matrix[i][j] = 0; } } for(int j = 1; j < n; j++) { if(matrix[0][j] == 0) { for(int i = 1; i < m; i++) matrix[i][j] = 0; } } if(colHasZero) { for(int i = 0; i < m; i++) matrix[i][0] = 0; } if(rowHasZero) { for(int j = 0; j < n; j++) matrix[0][j] = 0; } } }