LeetCode 73. Set Matrix Zeroes

原题链接在这里：https://leetcode.com/problems/set-matrix-zeroes/

题目：

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?

题解：

观察到判断一个数是否应该更改为0 只需要看它的同行同列是否有0就好，所以可以优化成只使用两个 boolean array, 一个记录所有行， 一个记录所有列 是否有0即可, 如此使用O(m+n) space.

更好的方法是使用第一行 和 第一列 来代替额上面的两个记录array, 那么第一行和第一列本来是否需要改成0 呢，其实只需要两个boolean value: firstRow, firstCol 来记录就好，如此只需要使用O(1) space.

Time Complexity: O(m*n).

Space: O(1).

AC Java:

public class Solution {
    public void setZeroes(int[][] matrix) {
        if(matrix == null || matrix.length == 0){
            return;
        }
        boolean firstRow = false;
        boolean firstCol = false;
        int m = matrix.length; 
        int n = matrix[0].length;
        
        //Record if first row, first columna contains 0
        for(int j = 0; j<n; j++){
            if(matrix[0][j] == 0){
                firstRow = true;
                break;
            }
        }
        for(int i = 0; i<m; i++){
            if(matrix[i][0] == 0){
                firstCol = true;
                break;
            }
        }
        
        //first scan, if there are 0, mark its corresponding row and columna as 0 at first column and first row
        for(int i = 1; i<m; i++){
            for(int j = 1; j<n; j++){
                if(matrix[i][j] == 0){
                    matrix[0][j] = 0;
                    matrix[i][0] = 0;
                }
            }
        }
        
        //Scan the matrix for the second time, if corresponding mark is 0, change this element to 0
        for(int i = 1; i<m; i++){
            for(int j = 1; j<n; j++){
                if(matrix[i][0] == 0 || matrix[0][j] == 0){
                    matrix[i][j] = 0;
                }
            }
        }
        
        //Change first row and columna at last
        if(firstRow){
            for(int j = 0; j<n; j++){
                matrix[0][j] = 0;
            }
        }
        if(firstCol){
            for(int i = 0; i<m; i++){
                matrix[i][0] = 0;
            }
        }
    }
}

类似Game of Life.