[leetcode][73] Set Matrix Zeroes
阿新 • • 發佈:2018-09-22
ons bad put integer 是否 int solution dev set
73. Set Matrix Zeroes
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?
解析
把矩陣中含有0的那一行和列全部置0.
參考答案
自己寫的:
class Solution { public void setZeroes(int[][] matrix) { Set<Integer> rows = new HashSet<>(); Set<Integer> cols = new HashSet<>(); for (int i = 0; i < matrix.length; i++) { for (int j = 0; j < matrix[i].length; j++) { if (matrix[i][j] == 0) { rows.add(i); cols.add(j); } } } for (Integer i : rows) { for (int j = 0; j < matrix[i].length; j++) { matrix[i][j] = 0; } } for (Integer i : cols) { for (int j = 0; j < matrix.length; j++) { matrix[j][i] = 0; } } } }
一開始沒註意題目的意思,這個是一般的解法,題目的意思是不要用額外的空間。
別人寫的:
public class Solution { 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; } } } }
把第0行和第0列留出來記錄那些行和哪些列需要置0,用fr和fc來記錄第0行和第0列是否需要置零。
[leetcode][73] Set Matrix Zeroes