Leetcode 073 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.

click to show follow up.

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?

 

1.O(m+n)space,两个数组m和n分别存储第i行和第j列有没有零，比较慢

2.constant space ,利用矩阵的第零行和第零列分别存储第i行和第j列有没有零，需要注意的是再用两个变量row和column来表示第零行和第零列是否有零存在，因为这是边界情况

public class S073 {
    public void setZeroes(int[][] matrix) {
        //O(m+n) space slow
/*        int [] m = new int[matrix.length];
        int [] n = new int[matrix[0].length];
        for (int i = 0;i<matrix.length;i++) {
            for (int j = 0;j<matrix[0].length;j++) {
                if (matrix[i][j] == 0) {
                    m[i] = -1;
                    n[j] = -1;
                }
            }
        }
        for (int i = 0;i<matrix.length;i++) {
            if (m[i] == -1) {
                for (int j = 0;j<matrix[0].length;j++) {
                    matrix[i][j] = 0;
                    if (n[j] == -1) {
                        for (int k = 0;k<matrix.length;k++) {
                            matrix[k][j] = 0;
                        }
                    }
                }
            }
        }  */
        //利用矩阵的第零行和第零列分别来存每一列和每一行是否有零存在
        int row = -1;
        int column = -1;
        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 ) {
                        row = 0;
                    } 
                    if (j == 0) {
                        column = 0;
                    } 
                    matrix[0][j] = 0; //第零行
                    matrix[i][0] = 0; //第零列
                }
            }
        }
        for (int i = 1;i<matrix.length;i++) {
            if (matrix[i][0] == 0) {
                for (int j = 1;j<matrix[0].length;j++) {
                    matrix[i][j] = 0;
                }
            }
        } 
        for (int j = 1;j<matrix[0].length;j++) {
            if (matrix[0][j] == 0) {
                for (int i = 1;i<matrix.length;i++) {
                    matrix[i][j] = 0;
                }
            }
        } 
        if (row == 0) {
            for (int i = 0;i<matrix[0].length;i++) {
                matrix[0][i] = 0;
            }
        } 
        if (column == 0) {
            for (int i = 0;i<matrix.length;i++) {
                matrix[i][0] = 0;
            }
        }
    }
}