LeetCode 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.

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.
Could you devise a constant space solution?

一開始想到了空間O(m + n)的算法，用一個數組arr[m+n]記錄某行或者某列是否需要設為0，遍歷一遍數組，若num[i][j]==0，則對應的賦值arr[i]=0,arr[m+j]=0即可。

沒想到constant的方法，上網搜一番之后發現：
沒有必要新開一個arr[m+n]數組，其實遍歷num[i][j]==0后，直接將num[i][0]與num[0][j]設為0即可~~確實啊！！！

不過對于首行和首列，這樣直接改變它們的值會影響它們原本的值，而首行與首列是否set為0又是由它們原本的值決定的，因此一開始先掃一遍首行與首列，確定它們是否需要set，再掃描i=[2...m]與j=[2...n]，判斷其他行與列。

代碼：

    public void setZeroes(int[][] matrix) {
        if (matrix.length == 0) return;
        int m = matrix.length, n = matrix[0].length;
        boolean row = false, col = false;
        
        // Check first row & col
        if (matrix[0][0] == 0) {
            row = true;
            col = true;
        } else {
            for (int i = 1; i < m; i++) {
                if (matrix[i][0] == 0) {
                    col = true; 
                    matrix[0][0] = 0;
                    break;
                } 
            }
            
            for (int j = 1; j < n; j++) {
                if (matrix[0][j] == 0) {
                    row = true; 
                    matrix[0][0] = 0;
                    break;
                }             
            }
        }
        
        // Check other rows & cols
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (matrix[i][j] == 0) {
                    matrix[i][0] = 0;
                    matrix[0][j] = 0;
                }
            }
        }
        
        // Assign rows
        for (int i = 1; i < m; i++) {
            if (matrix[i][0] == 0) {
                for (int j = 1; j < n; j++) matrix[i][j] = 0;
            }
        }
        
        // Assign cols
        for (int j = 1; j < n; j++) {
            if (matrix[0][j] == 0) {
                for (int i = 1; i < m; i++) matrix[i][j] = 0;
            }
        }
        
        // Assign first row & col
        if (col) {
            for (int i = 1; i < m; i++) matrix[i][0] = 0;
        }
        if (row) {
            for (int j = 1; j < n; j++) matrix[0][j] = 0;
        }
        
    }